Yes:
Assuming that $H_1$ has an element $u$ of order 2 and $\max(|H_1|,|H_2|)\ge 3$, there exist hyperbolic $h_1,h_2$ such that $[h_1,h_2]$ is conjugate to its inverse and nontrivial (and hyperbolic).
First case: $|H_2|\ge 3$. Pick two distinct nontrivial commuting elements $b,c$ in $H_2$ (exercise: these exist). Then writing $b^{-1}=B$, $c^{-1}=C$
$$b[ub,uc]B = b(BuCuubuc)B = uCbucB = uCbuBc,$$
which is hyperbolic and conjugate to its inverse $CbuBcu$ (it is indeed product of two elements of order 2: $u$ and $CbuBc$).
Second case: $|H_1|\ge 3$. Pick $a\in H_1\smallsetminus\{1,u\}$ and $b\in H_2\smallsetminus\{1\}$. Then, writing $a^{-1}=A$, $b^{-1}=B$
$$b[aub,ab]B=b(BuABAaubab)B=uABuba,$$
which is hyperbolic and conjugate to its inverse $ABubau$ (it is indeed the product of two elements of order 2: $u$ and $ABuba$).
If your claim (that the only possibility for a nontrivial commutator of hyperbolic to be conjugate to its inverse is to be product of two elements of order 2) is correct ($\ast$), then it is a necessary condition that $H_1$ or $H_2$ has an element of order 2.
($\ast$) I finally can check it (this thus addresses Neil Strickland's comment): suppose that some element of a free product $H_1 \ast H_2$ is hyperbolic and conjugate to its inverse. Let $P$ be the stabilizer of its axis in the Bass-Serre tree. Then $P$ acts on this axis and does not preserve an orientation. Let $Q$ be the fixator (=pointwise stabilizer) of this axis. Then $Q$ fixes a directed edge, which in the case of the Bass-Serre tree of a free product has a trivial stabilizer; Hence $Q$ is trivial. So $P$ acts faithfully on the axis, and hence is an infinite dihedral group. This proves that there exist elements of order 2, and more precisely shows that hyperbolic elements in $P$ are products of two elements of order 2.