We claim the answer is $Q=22$. Note that an arbitrary line $\ell$ will divide a circle into at most two parts. As these parts are continuous and connected, one is red and other blue. Thus, assuming not all colors are the same, we define a bounding pair $(i,j)$ to be integers $1\leq i,j\leq n$ such that $A_i,A_j$ are red but $A_{i-1},A_{j+1}$ are blue.

Corollary. For a point $P$ and $\mathcal R,\mathcal B$ with at least one point, the set of bounding pairs of $\mathcal R,\mathcal B$ is partitioned into bounding pairs of $\mathcal R\cup P,\mathcal B$ and $\mathcal R,\mathcal B\cup P$. Note the above proof does change even if $\mathcal R=\emptyset$ (for Alice's part, where it is in at most one of the sets)! The corollary implies that at least one of the sets contain half the number of valid bounding pairs, so any move Bob makes can not reduce the number of bounding pairs by more than a half. Thus, as there are $2022^2$ bounding pairs, Alice can force Bob to take at least \[\left\lceil\log_2 2022^2\right\rceil=22\]moves. We now show a strategy for Bob. Claim 4. If $\mathcal R,\mathcal B$ have at least one point each (and a total of 3 points in minimum), and $S$ is the set of valid bounding pairs, then there exists a point $P$ such that the valid bounding pairs of both $\mathcal R\cup P,\mathcal B\cup P$ is $\left\lceil\frac{|S|}2\right\rceil$. Proof. Assume $|\mathcal B|\geq 2$. Then, choose a point $X$ on $\mathcal R$, and suppose $B_1,B_2\in\mathcal B$. We can choose a point $Y$ on $B_1B_2$ such that $XY$ does not pass through any intersection of bounding pairs (by Pigeonhole; finitely many intersections, infinitely many lines). Move $P$ from $X$ to $Y$; then, the number of valid bounding pairs in $\mathcal R\cup P$ goes from $|S|$ to $0$, but decreases by at most 1. Thus, at some point it is our desired number, and from Claim 3, we are done (for the other set). Now, Bob just applies a binary search until he can find two points of opposite color. If he can't, he ends up with one of the situations that doesn't lead to a bounding pair. Each of these steps makes sure he does not decrease by more than $\frac 12$ (up to a $+\frac 12$), which implies he can finish in $22$ moves.

Quote: Claim 3. Assume $\mathcal R,\mathcal B$ are nonempty, and let $S$ be the set of possible bounding pairs. For any point $P$ added and $(i,j)$ in $S$, $A_iA_j$ is a valid pair for exactly one choice of color for $\mathcal R$ and $\mathcal B$. [asy][asy] size(10cm); import geometry; filldraw((-5,3)--(-3,2)--(0,0)--(-4,-5)--cycle,red*0.1+white,invisible); filldraw((6,4)--(2,2)--(1,0)--(3,-4)--cycle,blue*0.1+white,invisible); draw((-5,3)--(-3,2)--(0,0)--(-4,-5),red); draw((6,4)--(2,2)--(1,0)--(3,-4),blue); draw((-5,3)--(0.5,2.5)--(0,0),red+dashed); draw((6,4)--(0.5,2.5)--(1,0),blue+dashed); dot((0.5,2.5)); label("$P$",(0.5,2.5),N); label("$\mathcal R$",(-3,0),red); label("$\mathcal B$",(3,0.5),blue); draw(line((0.5,0),(-0.25,2)),darkgreen+dotted); label("line $A_iA_j$", (2,-4),W); [/asy][/asy] Proof. Let $(i,j)\in S$ be a bounding pair, and note that $A_iA_j$ intersects neither of $\mathcal R,\mathcal B$. We want to show the line $A_iA_j$ intersects at exactly one of $\mathcal R\cup P,\mathcal B\cup P$. Assume $R_1PR_2$ and $B_1PB_2$ are the points consecutive to $P$ in the new polygons. We know $P$ is inside the convex polygon $R_1R_2B_1B_2$, so thus any line must intersect its interior. This implies that to reach $B_2R_2$ from $B_1R_1$, it crosses at least one of $B_1P,R_1P$, implying it intersects at least one of the new polygons. We show now it intersects at most one polygon. We know that if $A_iA_j$ intersects both polygons, it intersects both triangles (as it is disjoint from $\mathcal R,\mathcal B$). But, this means that it intersects $R_1P,R_2P,B_1P,B_2P$ but not $R_1R_2,B_1B_2$. However, as $\mathcal R$ is contained between the lines $R_1P,R_2P$ (and same with $\mathcal B$). Thus, orient $A_iA_j$ horizontally, and suppose $\mathcal R$ is above $A_iA_j$. This implies $P$ is below $A_iA_j$, so $\mathcal B$ is above $A_iA_j$, implying $A_iA_j$ was not a bounding pair, a contradiction. Thus, $A_iA_j$ intersects at most one of $\mathcal R\cup P,\mathcal B\cup P$, implying the result. [asy][asy] size(10cm); import geometry; filldraw((-1,5)--(0,3)--(-4,1)--(-6,3)--cycle,red*0.1+white,invisible); filldraw((2,6)--(1,4)--(3,0)--(6,2)--cycle,blue*0.1+white,invisible); draw((-1,5)--(0,3)--(-4,1)--(-6,3),red); draw((2,6)--(1,4)--(3,0)--(6,2),blue); draw((0,3)--(0.5,-1)--(-4,1),red+dashed); draw((1,4)--(0.5,-1)--(3,0),blue+dashed); dot((0.5,-1)); label("$P$",(0.5,-1),S); label("$\mathcal R$",(-3.25,3),red); label("$\mathcal B$",(3,3),blue); draw(line((0,0),(1,0)),darkgreen+dotted); label("line $A_iA_j$", (-4,0),S); [/asy][/asy] But for a bounding pair there could be infinite many choices of the line ,not necessarily the line A_iA_j so there may be cases that $\mathcal R,\mathcal B\cup P$is valid for one line and$\mathcal R\cup P,\mathcal B$ valid for another line.

The answer is $22$ Proof of $Q\leq 22$. First, Bob plays on two diametrically opposite points on the $2022$ gon. Also, note that the blue and red points along the circle are a continuous string on points. We now split into cases. Case 1: The two points are different colors. Then, we repeatedly take the middlest vertex of $A$ on the top arc. This will create $2$ new arcs of the circle which we now the endpoints of. Take the arc which has different colored endpoints which is guaranteed to exist next and repeat. This is a binary search, taking $\lceil \log_2 1011\lceil=10$ moves on each arc, and since there are $2$ arcs it takes at most $2+10+10$ moves. Case 2: Both endpoints are the same color, say blue. Now, note that this forms two arcs, and since the reds are continuous, at most one of these arcs has red. Claim: Suppose we have points on $A$ named $P_1$, $P_2$, and $M$ all of which are blue. Then, in one search we can rule out the possibility of one of arcs $\widehat{P_1M}$ or $\widehat{MP_2}$ having a red point. Proof. Let the tangents to the circumcircle of $A$ at $M,P_1$ to meet at $K$. We claim that the search at $K$ works. Indeed, if $K$ is blue then $\triangle P_1MK$ is all blue, and we can see this implies it has blue endpoints. [asy][asy] import graph; size(14.6654837714418356cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.8583881841513751,xmax=0.8070955872904605,ymin=0.40117762796371864,ymax=1.2; pen ffwwzz=rgb(1.,0.4,0.6); pair A=(1.,0.), P_2=(0.5,0.8660254037844387), M=(-0.1736481776669302,0.9848077530122082), P_1=(-0.766044443118978,0.6427876096865397), K=(-0.5320888862379557,0.9216049851068763), D=(-0.5,0.8660254037844383), F=(0.766044443118978,0.6427876096865396), G=(-0.9396926207859071,0.34202014332567215), H=(0.9396926207859082,0.3420201433256689); filldraw(P_1--K--M--cycle,blue+opacity(0.10000000149011612),linewidth(1.)+blue); draw((-0.9396926207859085,0.34202014332566916)--P_1,linewidth(1.)); draw(P_1--(-0.5,0.8660254037844389),linewidth(1.)); draw((-0.5,0.8660254037844389)--M,linewidth(1.)); draw(M--(0.17364817766693047,0.9848077530122081),linewidth(1.)); draw((0.17364817766693047,0.9848077530122081)--P_2,linewidth(1.)); draw(P_2--(0.7660444431189781,0.6427876096865394),linewidth(1.)); draw((0.7660444431189781,0.6427876096865394)--(0.9396926207859084,0.3420201433256687),linewidth(1.)); draw(circle((0.,0.),1.),linewidth(1.)+linetype("4 4")+ffwwzz); draw(P_1--K,linewidth(1.)); draw(K--M,linewidth(1.)); draw(P_1--K,linewidth(1.)+blue); draw(K--M,linewidth(1.)+blue); draw(M--P_1,linewidth(1.)+blue); draw((-1.,0.)--G,linewidth(1.)); draw(H--A,linewidth(1.)); dot(A,ds);dot(K,ds) ; dot(P_2,ds); label("$P_2$",(0.5036155586852523,0.8755936967772827),NE*lsf); dot(M,ds); label("$M$",(-0.1700735409594419,0.9935121416268968),NE*lsf); dot(P_1,ds); label("$P_1$",(-0.7980578170190145,0.6507259647384837),NE*lsf); dot((-1.,0.),ds); label("$K$",(-0.5283993578667963,0.9286112921360239),NE*lsf); dot(D,linewidth(4.pt)+ds); dot((0.17364817766692964,0.9848077530122082),linewidth(4.pt)+ds); dot(F,linewidth(4.pt)+ds); dot(G,linewidth(4.pt)+ds); ; dot(H,linewidth(4.pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now, if $K$ is red, then we claim that $\widehat{MP_2}$ can't have a red point. FTSOC assume it did, call this point $P_3$. Then, consider $X=P_1M\cap KP_3$; we have that $K,P_3$ are red so $X$ is red as it is in the interior of that segment. But then a red point $X$ is inside a blue segment $P_1M$, contradiction. The claim is proved. $\blacksquare$ This allows us to narrow our search for red points by half. The strategy is then to use a move described in the claim to determine which half of the circle has a red point. Then, find the color of the midpoint of that arc. If it is blue, repeat. Else, it is red so do the binary search like above. The binary search as described will take $2\lceil \log_2 x \rceil$ where we are searching $x$ points. The process of using $2$ moves to narrow our search for red similarly takes $2\lceil \log_2x\rceil$ moves, implying it is possible in $2\lceil \log_2 2022\rceil$ moves. Proof of $Q\geq 2022$. I hate this. There are $c+2022\cdot 2021$ possible arrangements of which points of $A$ are red where $c<1434$. Since with $21$ moves there are $2^{21}$ possible results and $2^{21}<2022\cdot 2021$ we conclude that some coloring must correspond to $2$ colorings of $A$. This finishes.