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. 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.

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. 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. (B. von Kérékjártó, Über die endlichen topologischen Gruppen der Kugelfläche, Nederl. Akad. Wetensch. Proc. Ser. A vol. 22 (1919) pp. 568-569: There is probably a more accessible reference for this result, but I do not have one.) 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.

A side note: If you are willing to replace the symmetric space $X$ with the $n$-fold product $X^n=X\times ... \times X$, then every isometric action of $\Gamma$ extends to an isometric action of $\tilde\Gamma$ on $X^n$ (where $|\tilde\Gamma:\Gamma|=n$): This is the induced representation construction. Incidentally, the following is an open problem:

Let $\Gamma$ be a discrete isometry group of hyperbolic $n$-space $H^n$ and $\Gamma \subset \tilde\Gamma$ is a finite-index extension. Is it true that $\tilde\Gamma$ admits a properly discontinuous isometric action on $H^m$ for some $m$?

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. 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  $G=PO(3,1)$.

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. 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.