For the official version... (well, the two versions are more or less the same). Define $f:\mathbb{R}\to\{R,B\}$ so that $f(x)$ is the color of strip $S_{\lfloor{x}\rfloor}$. Let $ABCD$ be a free rectangle with sides $a=AB$ and $b=BC$ where $1\le a<b$, and assume for the sake of contradiction that we can not place it anywhere such that $f(A_x)=f(B_x)=f(C_x)=f(D_x)$. Considering the horizontal and vertical orientations, $f(i)\ne f(i+a)$ and $f(i)\ne f(i+b)$ for all $i$. Thus $f(i)=f(i+2a)=f(i+2b)$ as well, so if $a=gc$ and $b=gd$ where $g=\gcd(a,b)$ then $f(i)=f(i+2g)$. If one of $c,d$ is even, say $c$, then $f(i)=f(i+2g(c/2))=f(i+a)$, so $c$ and $d$ are both odd. In particular, $d\ge c+2\ge3$. Now set $A_x=i\in\mathbb{Z}$, $B_x-i=u=i-B'_x$, $D_x-i=v=i-D'_x$, and $C_x-i=u+v=i-C'_x$ (the $y$-coordinates don't matter). If we take $v=2g<b$, then by similar triangles $u=(cg/d)\sqrt{d^2-4}\not\in\mathbb{Z}$ (since $d\ge3$, $d^2-4$ cannot be a perfect square). Notice that \[f(A_x)=f(i)=f(i\pm2g)=f(i\pm v)=f(D_x)=f(D'_x).\]But we also have \[f(B_x)=f(i+u)=f(i+u+2g)=f(i+u+v)=f(C_x)\wedge f(B'_x)=f(C'_x),\]so by our assumption applied to $ABCD$ and $AB'C'D'$, \[f(i)=f(A_x)\ne f(B_x)=f(i+u)=f(i+\lfloor{u}\rfloor)\wedge f(i)\ne f(i+\lfloor{-u}\rfloor)\]and thus \[f(i+\lfloor{u}\rfloor)=f(i+\lfloor{-u}\rfloor)\]for all integers $i$. Yet $u$ is non-integral (in fact, irrational), so $\lfloor{u}\rfloor+\lfloor{-u}\rfloor=-1$ and $\gcd(\lfloor{u}\rfloor,\lfloor{-u}\rfloor)=1$. Furthermore, since $\lfloor{u}\rfloor$ and $\lfloor{-u}\rfloor$ have opposite parities, we conclude that $f(i)=f(i+1)$ for all $i\in\mathbb{Z}$, contradicting the fact that $f(i)\ne f(i+a)$ for all $i$.

I hope the following works: For sake of contradiction assume that no such rectangle with sidelenghts a,b respectively exists. First of all the color of a point only depent on its x-coordinate, so we don't have to care about the x-coordinates. I only consider rectangles which sides are parallel to the coordinate axis and wlog (by moving the rectangle in the y-direction) one side lies on the x-axis. By "x is colored white" I simply mean that all points with x-coordinate x are colored white. Consider the set S of points with x-coordinate $b-x$ as x varies from 0 to b, 0 excluded. Assume wlog that $b-t$ is colored white. Then -t has to be colored black, if not because of $|b-t-(-t)|=b$ the rectangle enclosed by the lines $x=-t, x=b-t, y=0, y=a$ would give a valid rectangle. Now by definition all the points in the interval $[-\left\lceil{t}\right\rceil,-\left\lfloor{t}\right\rfloor[$ are of the same color and thus they are colored black. Now take $u\neq t$, $u \in ]0,b]$. If $b-u$ is colored black the same argument as before yields that all points in the interval $[-\left\lceil{u}\right\rceil,-\left\lfloor{u}\right\rfloor[$ are of the same color, i.e. white. Since this holds for all $u,t \in ]0,b]; u \neq t$ choose u,t s.t. $m<u,t<m+1$ for some $m \in \mathbb N$ and $u,t \in ]0,b]$. Then by definition it follows that $\left\lceil{u}\right\rceil=\left\lceil{t}\right\rceil=m+1$ and $\left\lfloor{u}\right\rfloor=\left\lfloor{t}\right\rfloor=m$. This implies that all the points in the interval $[-m-1,-m[$ are colored both black and white. Contradiction. Thus all the points in S are colored in the same color, say wlog white. Replacing b with a and vice versa in the above arguments it likewise follows that all the points in the interval $[0,a[$ are colored in the same color. Now wlog take a>b. Then $[0,b]\in [0,a[$, so all the points in $[0,b]$ are colored with the same color. Now the vertices of the rectangle enclosed by the lines $x=0,x=b,y=0,y=a$ are of the same color, which finishes the proof. mathe760

Let $f(n)$ be defined as follows $f(n)=1$ if $S_{n}$ is colored black and $f(n)=-1$ otherwise. Assume that there is some rectangle $a x b$(WLOG $a>b$) where $a\not = b$ so that it cannot be placed on the board with all vertexes having the same color(which are later refered to as good retangles). Now if all $f(n)$ are equal that's a contradiction. Otherwise an integer $n$ exists such that $f(n-1)\not = f(n)$ then chose 2 points of the retangle to be $(0,n),(a,n)$ we easily get that if $b$ is not an integer we can move that rectangle left or right getting it to be a good rectangle a contradiction. Now do the same for $a$ So we have that both $a,b$ are integers. Now notice(with assumption that no $a x b$ rectangle is good) that $f(n)=-f(n+a)=f(n+a+b)$ so $f$ is periodic with period $a+b$ but $f(n+a)=-f(n)=f(n+b)$ so $f$ is periodic with period $a-b$ but then $f$ is periodic with $gcd(a-b,a+b)|2d$ (where $d=gcd(a,b)$) Now let $a=dx$ $b=dy$ since $f$ is not periodic with period $a$ then $x$ is odd. Now choose again the $n$ with $f(n-1)=-f(n)$ and choose points $A(n,0),B(n+(x-1)d,y)$ where $AB=a$ then it's not hard to prove that rectangle $ABCD$ which is $a x b$ has non integer $x$-coordinates for $C,D$ and we can move it left or right to get a good rectangle. The last just narrows down(after calculating the coordinates) to $x^2-1$ not being a square of an integer since $x>1$ because $a>b$.

If $a$ is not an integer then just pick a rectangle with sides parallel to the axes and move it a little bit, you get any 2 consecutive stripes have the same color => all the plane is the same color => done. Same with $b$, so $a,b$ are integers. Now any two stripes at distance $a$ or $b$ have distinct colors and by Bezout this implies $v_2(a)=v_2(b)$. By Bezout we get any two stripes at distance $d=gcd(a,b)$ have distinct color, so at distance $2d$ they have the same color. So if $a>b$, $2d<a$ and so we can assume two vertices of the rectangle at distance $a$ are at a vertical distance of $2g$, and the same happens for the two opposite vertices. Now if we shift this rectangle up horizontally, by the same reasoning as in the beginning, eventually all 4 vertices will have the same color. QED

Edit: Clear

Suppose the contrary, namely that there exists a rectangle $m\times n$ which can never be monochromatic. If $m$ is not an integer, then consider placing the rectangle with side $m$ parallel to the $x$ axis. Then, consider shifting it horizontally such that the left edge remains in the same strip. As $m$ is nonintegral, the right end traverses two strips, so these two strips must be the opposite color of the left strip, and hence the same color. Repeating this argument as the left edge varies over different strips, we get that all strips have the same color, which is obviously not good. Thus, WLOG $m,n$ are both integers. WLOG $m>n$. Considering only rectangles with sides parallel to the axes, we get that strips $m$ and $n$ apart have distinct colors. So, the colors have period $2\gcd(m,n)$, and strips $\gcd(m,n)$ apart have distinct colors. Note that this means $\nu_2(m)=\nu_2(n)$, since if $\nu_2(m)>\nu_2(n)$, we will have strips $m$ apart having the same color. In fact, we can say that "if strips $k$ apart have distinct colors," then $\nu_2(k)$ is fixed. Now, suppose we take the rectangle and tilt it such that the x coordinates of the bottom points are exactly $m-\gcd(m,n)$ away. Then, they have the same color, so we only need to consider the x coordinates of a single "vertical" edge. Suppose that the difference in $x$ coordinates of the image of a vertical edge is $d$. Obviously, $d$ is nonintegral, so if no monochromatic $m\times n$ exists, strips $\lfloor d\rfloor$ apart and $\lceil d\rceil$ apart have different colors. However, exactly one of these is even, which is contradiction to all $\nu_2$s being equal, as desired.

Solved with eisirrational, goodbear, Th3Numb3rThr33. Let this rectangle have dimensions \(a\times b\), and for each \(i\), let strip \(S_i\) have color \(c_i\). Claim: If \(a\not\in\mathbb Z\), then we are done. Proof. Assume for contradiction an \(a\times b\) rectangle may not be placed in the plane in such a way. Only consider axis-aligned rectangles with horizontal edge \(a\) and vertical edge \(b\). Then, \(c_n\ne c_{n+\left\lfloor a\right\rfloor}\), but \(c_n\ne c_{n+\left\lceil a\right\rceil}\); thus \(c_{n+\left\lfloor a\right\rfloor}=c_{n+\left\lceil a\right\rceil}\). As we vary \(n\), all strips have the same color, so we are clearly done. \(\blacksquare\) Henceforth \(a\), \(b\) are integers. The goal is to tilt the rectangle slightly in order to apply the above argument again. Again, assume for contradiction the problem is false. Claim: \(\nu_2(a)=\nu_2(b)\). Proof. Assume \(\nu_2(a)<\nu_2(b)\). Then we have \(c_0\ne c_a\ne c_{2a}\ne\cdots\ne c_{\operatorname{lcm}(a,b)}\); since \(\operatorname{lcm}(a,b)/a\) is even, \(c_0=c_{\operatorname{lcm}(a,b)}\). \(c_0\ne c_b\ne c_{2b}\ne\cdots\ne c_{\operatorname{lcm}(a,b)}\); since \(\operatorname{lcm}(a,b)/b\) is odd, \(c_0\ne c_{\operatorname{lcm}(a,b)}\). This is a clear contradiction. \(\blacksquare\) Claim: \(c_n\ne c_{n+\gcd(a,b)}\); in particular, \(c_n=c_{n+2\gcd(a,b)}\). Proof. Clearly \(c_n\ne c_{n+a}\) and \(c_n\ne c_{n+b}\) for all \(a\), \(b\). By B\'ezout's lemma we can select integers \(m\), \(n\) so that \(am+bn=\gcd(a,b)\). Then \(m+n\) is odd by parity, so \(c_n\) and \(c_{n+am+bn}\) have different colors. \(\blacksquare\) [asy][asy] size(4cm); defaultpen(fontsize(10pt)); pair A,B,C,D,WW,X,Y,Z; A=(0,5); B=(10,29); C=(22,24); D=(12,0); WW=(0,0); X=(0,29); Y=(22,29); Z=(22,0); draw(WW--X--Y--Z--cycle,gray); draw(A--B--C--D--cycle); label("\(a\)",B--C,unit(A-B)); label("\(b\)",D--C,unit(B-C)); label("\(t\)",D--Z,S); label("\(\frac ab\sqrt{b^2-t^2}\)",( (2B+Y)/3)--Y,N); dot("\(A\)",A,W); dot("\(B\)",B,NW); dot("\(C\)",C,E); dot("\(D\)",D,S); [/asy][/asy] Consider a tilted \(a\times b\) rectangle as shown above, with \(t=2\gcd(a,b)\). Then \(A\), \(B\) are in strips with the same color, and so are \(C\), \(D\), so \(B\), \(C\) must be in strips of different color. Let \(s:=\frac ab\sqrt{b^2-t^2}\) be the vertical distance between \(B\), \(C\). If \(s\in\mathbb Q\), then \(\sqrt{b^2-t^2}\in\mathbb Z\); however, \[\sqrt{b^2-t^2}=\gcd(a,b)\cdot\sqrt{\left(\frac b{\gcd(a,b)}\right)^2-4}\notin\mathbb Z,\]so \(s\) is not an integer. Then for each \(n\), \(c_n\ne c_{n+\left\lfloor s\right\rfloor}\), but \(c_n\ne c_{n+\left\lceil s\right\rceil}\); thus \(c_{n+\left\lfloor s\right\rfloor}=c_{n+\left\lceil s\right\rceil}\). As we vary \(n\), all strips have the same color, so we are done.

Let $c_n$ be $1$ if $S_n$ is red and $-1$ otherwise. Suppose there doesn't exist a rectangle with vertices same color. If $a$ was noninteger, then let $a'$ be the floor of $a$. Then, by setting the $a$-side parallel to the $y$-axis and its left coordinates on $x=0$, we have $c_0=-c_{a'}$. Doing the same but on $x\to 1_{-}$, we have $c_0=-c_{a'+1}$, so $c_{a'}=c_{a'+1}$. Thus, the sequence $c$ is constant, contradiction to $c_0=-c_{a'}$. Hence, both $a$ and $b$ are integers. Then, we can WLOG assume that $\text{gcd}(a,b)=1$, otherwise we can find $p\mid \text{gcd}(a,b)$ and simplify and strengthen the problem by simply merging all strips $S_k$ for $p\nmid k$ with the previous strip $S_l$ with $p\mid k$ and then dividing everything by $p$. Note that $c_0=-c_a$ so $c_{ab}=(-1)^a c_0=(-1)^b c_0$ so $a$ and $b$ are same parity. By the assumption, both are odd. Let $ar+bs=t$ for some integers $r$ and $s$, then \[t=ar+bs\equiv r+s\pmod 2\]so $c_0=c_t$ if and only $t$ is even. WLOG, let $c_{2k+1}=1$ and $c_{2k}=-1$ for all positive integers $k$. Now let's tilt the rectangle. The coordinates are: $A=(0,0)$, $B=(a,0)$, $D=(0,b)$, and $C=(a,b)$. If we slide $D$ right to $(2,b)$ while keeping the dimensions the same and while keeping $A$ on the $x$-axis, $A$ will end up at $(0,b-\sqrt{b^2-4})$. WLOG assume $b>a$ so $b\ge 3$. Then, note that $b^2-4> (b-1)^2$ so $\lfloor b-\sqrt{b^2-4} \rfloor=0$. Thus, $A$ and $D$ are both still blue. $B$'s $x$-coordinate will be $\sqrt{b^2-4}\cdot\tfrac{a}{b}$. But \[x = \sqrt{b^2-4}\cdot\frac{a}{b} > a\cdot \frac{b-1}{b}>a\cdot \frac{a-1}{a}=a-1\]so $B$ is blue. Since $D$'s $x$-coordinate is exactly two more than that, it is also blue. We are done.