Yes, here's an example with an HNN over finite index subgroups as requested. It's based on constructing an amalgam of two f.g. virtually free groups, that has no proper finite index subgroups, using Burger-Mozes groups.

Fact (proved below): *for every $n\ge 3$ there exists a non-torsion-free, virtually free group $G$ with a subgroup of finite index $H$, isomorphic to $F_n$, such that $G$ is normally generated by $H$.*

By Burger-Mozes, there exists $n<m$ with two embeddings $u,v$ of $F_m$ as finite index subgroup of $F_n$, such that the resulting amalgam $A(u,v)$ of $F_n$ and $F_n$ over $F_m$ using the embeddings $u,v$ is simple. (D. Rataggi improved and made explicit the values, providing for instance $(n,m)=(9,81)$.)

Now use $G$ as in the fact, with $H$ free of rank $n$ as above. Identify $H$ to $F_n$ to deduce two embeddings $u',v'$ from $F_m\to H\subset G$. Consider the amalgam $A=A(u',v')$ (of $G$ and $G$ over $F_m$ using $u'$ and $v'$).

Claim: *$A$ has no proper finite index subgroup.* Proof: the finite residual contains the Burger-Mozes subgroup given as subamalgam of $F_n$ and $F_n$ over $F_m$, hence, since $F_n$ normally generates $G$, contains both amalgamated factors $G$, hence is all of $G$. (Since $A$ is not torsion-free, it follows that $A$ is not virtually torsion-free.)

Consider the HNN extension $B$ given by the pair of embeddings $u',v'$ of $F_m$ into $G$. Namely, this is $(\langle t\rangle\ast G)/R$, with $R$ normally generated by the $tu'(h)t^{-1}v'(h)^{-1}$ for $h\in F_m$. Then $B$ is also not virtually torsion-free (the quotient by its finite residual is $\mathbf{Z}$), since it contains the above amalgam.

Proof of the fact: it is enough to do it for

$n=3$, yielding

$H\le G$: in general just use the projection

$G'=F_{n-3}\ast G\to G$ and define

$H'$ as the inverse image of

$H$.

Consider the virtually free group $G=\langle a,b:b^3=1\rangle$. Map it by $p$ onto the dihedral group $D_6$ mapping $a$ to an element of order 2 and $b$ to an element of order 3. Then the kernel $K$ of $p$ has index 6 and is free.

Now consider a subgroup $L$ of $G$ of index 3 containing $K$, this generating normally $G$. Since $L$ has a torsion-free subgroup of index 2, it has no element of order 3, and hence $L$ is free. Moreover $L$ has rank 3 (subfact below).$\square$

Subfact: *the free group $L$ has rank 3.* Proof:
Note that $G$ also has a normal subgroup $N$ of index 3, free of rank $3=1+2$ (namely the kernel of the retraction from $G$ to $\langle b\rangle$ killing $a$, which is freely generated by $a$, $bab^{-1}$, $b^2ab^{-2}$).
Since in a virtually free group all free subgroups of a given index have the same rank, we deduce that $L$ has rank 3.