I think it is torsion-free when $\mathcal{G}$ is torsion free or $Z=\text{Supp}(\mathcal{G})$ has codimension one. Please correct me if the argument has error.

We have an exact sequence
$$0\rightarrow \mathcal{H}/\mathcal{F}\rightarrow(\mathcal{O}_X\otimes V)/\mathcal{F}\rightarrow\mathcal{G}=(\mathcal{O}_X\otimes V)/\mathcal{H}\rightarrow0$$

When $\mathcal{G}$ is torsion-free, then any torsion subsheaf is mapped to zero in $\mathcal{G}$ and thus lies in $\mathcal{H}/\mathcal{F}$. It is impossible since $\mathcal{H}/\mathcal{F}$ is torsion-free.

Assume that $\mathcal{F}_0:=(\mathcal{O}_X\otimes V)/\mathcal{F}$ admits a non-trivial torsion sheaf say $\mathcal{T}$, then $\mathcal{T}$ is supported on $Z=\text{Supp}(\mathcal{G})$. When $Z\subset X$ has codimension $k>0$, then $c_i(\mathcal{T})=0$ for $i=0,1,\dots,k-1$. 

In particular, $c_1(\mathcal{T})=r[Z]$ for some $r>0$ when $k=1$. It will lead to $\mu(\mathcal{F}_0/\mathcal{T})>0$, which will contradict to the stability of $\mathcal{O}_X\otimes V$.