Let $G$ be a group (not necessarily countable discrete) which contains a free subgroup $F=F_2=\langle a, b\rangle$, denote $H=\langle a, b^2\rangle$, and assume that the centralizer of $F$ inside of $G$ is trivial, i.e., $C_G(F)=\{e\}$.
Suppose $e\neq g\in G$, and $gag^{-1},~ gb^2g^{-1}\in H$, is it true that $gbg^{-1}\in F$?
Any hint? Thanks in advance!
No. Just take an HNN-extension. For example, take $G$ as follows. $$G=\langle x, y, t; txt^{-1}=x, ty^2t^{-1}=y^2\rangle$$ Then $tyt^{-1}$ is a $t$-reduced word and so is not contained in $\langle x,y\rangle$ by Britton's lemma.
Note: the centraliser of $H=\langle x, y\rangle$ is trivial here, so this answers your question. However, the centraliser of $t$ is not trivial. If you want to change this then you could the relations I gave with the relations $txt^{-1}=x^2$ and $ty^2t^{-1}=y^4$, or some similar choices which don't correspond to finite-order element of $\operatorname{Out}(F_2)$ (or rather, don't restrict to such a finite order element on either $x$ or $y$).