Suppose $\phi : B \to B$ is a quasi-isometry of a neutered space $B$ (so $B$ is obtained by removing a collection of disjoint open horoballs from $\mathbb{H}^n$, and the metric $d_B$ on $B$ is the path metric induced by the hyperbolic metric $d_\mathbb{H}$). Is it possible to extend $\phi$ to a quasi-isometry on $\mathbb{H}^n$? That is, does there exist a quasi-isometry $\hat\phi : \mathbb{H}^n \to \mathbb{H}^n$ such that $\hat\phi \vert_B = \phi$?
I have some ideas for how to define $\hat\phi$ (assuming the answer to my question is yes), but I have some concerns which leads me to think that maybe the answer to my question is no. If such an extension exists, then it seems to imply that the quasi-isometry $\phi$ of $(B,d_B)$ is also a quasi-isometry (with probably different QI constants) of $(B,d_\mathbb{H})$. To my intuition however, this seems unlikely. While we do know that $d_\mathbb{H}(p,q) \le d_B(p,q)$ for points $p,q \in B$, I do not know of a sufficiently strong (e.g. coarse Lipschitz) inequality going the other direction. So I don't see much reason to believe that $\phi : (B,d_\mathbb{H}) \to (B,d_\mathbb{H})$ will be a quasi-isometry.
If I am right that $\phi$ cannot be directly extended, then is there any other natural way to construct a quasi-isometry of $\mathbb{H}^n$ from $\phi$?
Here's a way to see that it isn't always possible to build such an extension, at least in $\Bbb{H}^2$. Take a once-cusped finite volume hyperbolic surface $S$, and cut off the cusp with a horosphere, leaving a compact surface with boundary. The universal cover of this cut-off surface is a neutered space $X$ in $\Bbb{H}^2$, which is given the path metric $d_X$.
The fundamental group of $S$ is free, and it acts by cocompactly and by isometries on $(X, d_X)$. So Svarc-Milnor tells you that $(X, d_X)$ is quasi-isometric to a tree. The boundary at infinity of this tree is a Cantor set, which maps (non-injectively) into the ideal boundary of hyperbolic space.
Any quasi-isometry of $(X, d_X)$ which extends to a quasi-isometry of $\Bbb{H}^2$ has to preserve the cyclic order on the ideal boundary of $\Bbb{H}^2$. But, you can find plenty of quasi-isometries of trees that don't do this.
This argument is pretty special to $\Bbb{H}^2$. I am not actually sure what should happen in higher dimensions, since you may be able to argue in those cases that a quasi-isometry of the neutered space must permute the horospheres (although maybe not). [Edit: see YCor's comment below for the answer to this question.]