By the generalized word theorem, we know that if we have $H ≤ F_A$ finitely generated and $g \in F_A$, then $g \in H$ iff $\overline{g}$ is accepted in $S_H$ (Stalling's Automata of H).
Then, $xgx^{-1} \in H$ is simply saying that $\overline{x g x^{-1}}= \overline{x} \overline{g} \overline{x^{-1}}$ is accepted by $S_H$.
I'm not sure where to go from this ...
If $\overline{x g x^{-1}}$ is accepted by $S_H$, then $S_H$, must have a tail $\overline{x}$. If doesn't we can already say that $xgx^{-1} \not\in H$. If has a tail, then if we remove the tail, we get an automata that must accept $\overline{g}$.
Is that it?
$F_A$ is the free group with base $A$.
$\overline{x}$ is the reduced form of $x$.
Since we can replace $g$ by a conjugate, we can assume that $g$ is cyclically reduced, and that the reduced word for $xgx^{-1}$ is $wg'w^{-1}$ for some word $w$, where $g'$ is a cyclic conjugate of $g$.
If there is a repeated state when reading $w$ through the Stallings automaton of $H$, then we have $w=w_1w_2w_3$ where the states after reading $w_1$ and $w_1w_2$ are the same.
So $Hw_1 = Hw_1w_2$, and then $wg'w^{-1} \in H$ if and only if $w_1w_3g'(w_3w_1)^{-1} \in H$.
So we can assume that the length of $w$ is at most the number of states of the Stallings automaton for $H$, which reduces the problem to a finite number of tests of elements for membership of $H$.