Take $\tilde{\Gamma}$ to be a *higher genus Baumslag-Solitar group*, see section 9 of 
[M. Kapovich, B. Kleiner, Coarse Alexander duality and duality groups, Journal of Differential Geometry,  Vol. 69, (2005)  Number 2, p. 279-352](http://www.math.ucdavis.edu/~kapovich/EPR/ad.pdf). Group $\tilde\Gamma$ cannot act properly discontinuously on any contractible 3-manifold, but it contains a finite index subgroup $\Gamma$ which is is Gromov-hyperbolic and isomorphic to the fundamental group of a compact 3-manifold with boundary. Hence, by a corollary of Thurston's geometrization theorem, $\Gamma$ acts isometrically and discretely on hyperbolic 3-space, i.e., is a discrete subgroup of 
$G=PO(3,1)$. 

Edit: Here is an easier example. Note first that every finite group admitting an effective topological action on $R^2$ is either cyclic or dihedral. Now, take, say, a simple noncyclic  finite group $F$ and set $\tilde\Gamma= F *  F$. Then $\tilde\Gamma$ is virtually free, so it contains a free finite index subgroup $\Gamma$, which embeds in $PSL(2, R)$. On the other hand, every topological $\tilde\Gamma$-action on the plane has to be trivial since each free factor has to act trivially.