Is a HNN extension of a virtually torsion-free group virtually torsion-free?

Let $$G=\langle X\ |\ R\rangle$$ be a (finitely presented) virtually torsion-free group. Let $$H,K be isomorphic (finite index) subgroups of $$G$$ and let $$\varphi:H\rightarrow K$$ be an isomorphism.

Define the HNN extension $$\Gamma$$ of $$G$$ and $$\varphi$$ in the usual way, i.e. $$\Gamma=\langle G,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rangle$$.

Is $$\Gamma$$ virtually torsion-free?

My thought is that if $$T$$ is a finite index torsion-free subgroup of $$G$$ and if $$H$$ and $$K$$ are finite index we should be able to look at the intersection of each of them with $$T$$. So the group $$\langle T,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$$ would be a finite index torsion-free subgroup of $$\Gamma$$.

Also, can we say anything about the smallest index of a torsion-free subgroup? For example if $$G$$ contains a torsion-free subgroup of index $$k$$, does $$\Gamma$$ contain a torsion-free subgroup of index $$k$$? Or is the index bounded by some function of $$k$$?

No, even assuming that $$H,K$$ have finite index in $$G$$: see https://mathoverflow.net/a/330658/14094, where I provided an example (where $$G$$ is finitely generated, virtually free, and $$H,K$$ are free subgroups of the same finite index). It relies on Burger-Mozes groups.

If one does not assume that $$H,K$$ have finite index, it's even simpler (see for instance https://mathoverflow.net/a/330655/14094).