(This is not really an answer, rather a suggestion that the answer is probably negative.)

It is known that for linear groups over integers starting from $GL(4,\mathbb{Z})$ the membership problem is undecidable. Consider a finitely generated subgroup $G\subset GL(n,\mathbb{Z})$ with undecidable membership. It is, of course, very much decidable whether $g\in G$ is a square in $GL(n,\mathbb{Z})$ (using a Jordan normal form). There may be several square roots $h_1,h_2,\dots,h_k$, and the problem is to find out if any of them belong to $G$. I suspect that this restricted version of the membership problem is still undecidable in general, although I do not have a proof of this.

(This is not really an answer, rather a suggestion that the answer is probably negative.)

It is known that for linear groups over integers $GL(n,\mathbb{Z})$ starting from $n=4$ the membership problem is undecidable. (It is decidable for $n=2$. The case $n=3$ is an open problem for what I know.) Consider a finitely generated subgroup $G\subset GL(n,\mathbb{Z})$ with undecidable membership. It is, of course, very much decidable whether $g\in G$ is a square in $GL(n,\mathbb{Z})$ (using a Jordan normal form). There may be several square roots $h_1,h_2,\dots,h_k$, and the problem is to find out if any of them belong to $G$. I suspect that this restricted version of the membership problem is still undecidable in general, although I do not have a proof of this.