I was able to come up with the bound of $R\leq C\cdot N^{\frac{-m+2}{2}}$. Hopefully someone can sharpen the bound.

Solved with Rohan and Atul (mostly them). At its heart, this is a finite differences problem. Let $d=m-1$ for convenience, and say we are picking positive integers $a_0,a_1, \dots, a_d$ instead. We will assume $d$ is fixed from now on whenever we use asymptotic notation. Assume $N$ is sufficiently large, say $N>(d+1)^{1000}1000^{d+1}$ (the smaller $N$ can be taken care of by the $C(m)$). Let $n$ be the largest positive integer such that $4^d(n+d)<N$; note that $n>d$ and $n=\Omega(N)$. Now let $$a_i=\binom{d}{i}^2(n+i)<4^d(n+i)<N.$$Take alternating sum of square roots of $a_i$, that is, $$r=\sqrt n \sum_{i=0}^d (-1)^i \binom{d}{i} \sqrt{1+\frac{i}{n}}.$$Note that $\sqrt{1+x}$ has a power series expansion around $0$ when $|x|<1$; say $\sqrt{1+x}=b_0+b_1x+\cdots$. We in fact know a closed form from binomial theorem: $b_0=1$ and $b_i=\frac2{i}\binom{2i-2}{i-1}\left(-\frac14\right)^{i}$ for $i \geq 1$, but all we need is that $b_i \neq 0$ for all $i$. Now, expand every term of $r$ as the above Taylor series: $$r=\sqrt n \sum_{j=0}^\infty \frac{b_j}{n^j} \left(\sum_{i=0}^d (-1)^i \binom{d}{i} i^j \right).$$The $j^{th}$ term is the $d^{th}$ finite difference of the polynomial $x^j$ (starting from $0$), so it is $0$ for $j<d$, $d!$ for $j=d$, and $O(j^dd^j)$ for $j>d$ (here the constant is independent of $j$). Hence $$r=b_d d! \cdot n^{-d+\frac12}+\sqrt n \cdot O\left(\sum_{j=d+1}^\infty \frac{b_j j^d d^j}{n^j} \right) = b_d d! \cdot n^{-d+\frac12}+O(n^{-d-\frac12}).$$Since $b_d d! \neq 0$, for sufficiently large $n$, $r \neq 0$. If $r<0$, replace $r$ by $-r$. In conclusion, we get $$0<R \leq r=O(n^{-d+\frac12}) = O(N^{-d+\frac12}) = O(N^{-m+\frac32}).$$

Effectively the same as the above solution. The motivation is the $m=2$ case: we must take a difference $\sqrt{x}-\sqrt{y}$, and the only tool we have is that $\sqrt{N}-\sqrt{N-1}=O(N^{-\frac12})$. We are trying to spam finite differences to make things small. As an example for $m=4$, we take \begin{align*}\Big(\big(\sqrt{x}-\sqrt{x+1}\big)&-\big(\sqrt{x+1}-\sqrt{x+2}\big)\Big)-\Big(\big(\sqrt{x+1}-\sqrt{x+2}\big)-\big(\sqrt{x+2}-\sqrt{x+3}\big)\Big)\\ &=\sqrt{x}-3\sqrt{x+1}+3\sqrt{x+2}-\sqrt{x+3}\\ &=\sqrt{x}-\sqrt{9x+9}+\sqrt{9x+18}-\sqrt{x+3} \end{align*} The term under the root is $\binom{m-1}{i}^2(x+i)\leq\left(\max_{0\leq i\leq m-1}\binom{m-1}{i}\right)^2(x+m-1)$. So $x$ can be chosen as a linear function of $N$ such that all terms are $<N$. Since this finite difference is $O(x^{-(m-1)+\frac12})$ as proven above (I'm not going to repeat calculations), and $x=\Theta(N)$, we get the absolute value of sum of roots is $O(N^{-m+\frac32})$ as desired. (Note that we also need to ensure that the finite difference is nonzero. To avoid further calculations, we can cheekily force $x$ to be $p\pmod{p^2}$ for some prime $p>m$, so $x$ is the only term under the root that is a multiple of $p$ and not divisible by $p^2$. This ensures that the sum cannot be an integer and hence is nonzero.) Is it possible to improve the exponent of $-m+\frac32$? I'm pretty sure the finite difference method should get at best $\Theta(N^{-m+\frac32})$, so other methods must be considered...

Well, considering the improvement; Let $n$ and $k$ be a positive integers and $a_i,b_i\in \{1,\cdots ,n\}$, denote by $r(n,k)$ the minimum positive value of \[|\sum^k_{i=1}{\sqrt{a_i}}-\sum^k_{i=1}{\sqrt{b_i}}|.\]We shall prove $r(n,k)=O(n^{-2k+\frac{3}{2}}),$ and this bound can be obtained by putting $a_i={\binom{2k-1}{2i}}^2(n+2i),\ b_i={\binom{2k-1}{2i+1}}^2(n+2i+1)$. Notice that we need also that ${\binom{2k-1}{2i}}^2(n+2i),\ {\binom{2k-1}{2i+1}}^2(n+2i+1)<N$.