$\DeclareMathOperator\Syl{Syl}$$G$ is a finite solvable group and $G=PQR$, where $P\in \Syl_{p}(G)$, $Q\in \Syl_{q}(G)$, $R\in \Syl_{2}(G)$ and $|R|=2$. Suppose that $C_P(R)=P$ and $C_Q(R)=1$.
Since $|R|=2$, we have that $N_G(R)=C_G(R)=PR$.
Set $R^G=\langle R^g\mid g\in G\rangle$. Can we get that $R^G\leq QR$ ?
I think the answer is yes. I will assume that $p,q,r$ are all distinct. (We could have $p=r=2$ but then we get $P=R$ and $G=QR$ so the answer is still yes.) So $PQ$ is a normal odd order subgroup of index $2$ in $G$.
Proof by induction on $|G|$. Base cases when $P=1$ or $Q=1$ are easy so assume they are both nontrivial. If there is a nontrivial normal $q$-subgroup $N$ of $G$, then the result follows by applying the inductive hypothesis to $G/N$.
If not then we must have $O_p(G)\ne 1$. Then $C_G(O_p(G))$ is normal in $G$ and contains $R$, so $R^G \le C_G(O_p(G))$, and we may assume that $G=C_G(O_p(G))$.
But then $G/O_p(G)$ either has order $2$ and the result follows, or it has a nontrivial minimal normal subgroup $q$-subgroup $N/O_p(G)$, and a Sylow $q$-subgroup of $N$ is normal in $G$ and we are back to the first case.
\DeclareMathOperatorin math mode when appropriate rather than leaving math mode, e.g., for $\DeclareMathOperator\Syl{Syl}P \in \Syl_p(G)$$\DeclareMathOperator\Syl{Syl}P \in \Syl_p(G)$vs. $P \in$ Syl$_p(G)$$P \in$ Syl$_p(G)$. The latter has unpleasant side effects, e.g., the fonts don't match, and it allows a line break betweenSyland the subscript. (The\DeclareMathOperatorneed be done only once, and can then be used throughout the post.) I have edited accordingly. $\endgroup$