Here is a simple counter-example. Let $p : S_3 \to S_2$ be a degree 2 covering map from the closed genus 3 surface to the closed genus 2 surface, inducing an index 2 injection $p_* : \pi_1(S_3) \to \pi_1(S_2)$. Pulling back any hyperbolic structure from $S_2$ to $S_3$ via $p$ defines an embedding of Teichmuller spaces $p^* : T(S_2) \to T(S_3)$, the domain having dimension 6 and the range having dimension 12. Pick a point in $T(S_3)$ which is not in the image of this embedding nor in any translate of the image under the action of the mapping class group $MCG(S_3)$ on $T(S_3)$. This point represents a discrete subgroup $\Gamma \approx \pi_1(S_3)$ of $Isom(H^2) = PSL(2,R)$ such that, under the index 2 inclusion of $\Gamma \hookrightarrow \tilde\Gamma \approx \pi_1(S_2)$, the action of $\Gamma$ does not extend to an action of $\tilde\Gamma$.
ADDED AFTER INITIAL COMMENTS: The OP has clarified that he intends only to ask about the existence of topological actions of $\tilde\Gamma$, in which setting this is not a counter-example as pointed out by Yves.
However, in the other direction, rigidity theorems apply to give many interesting positive examples. For instance, suppose $\Gamma$ is a uniform lattice in $Isom(H^n) = SO(n,1)$, $n \ge 3$, meaning a discrete cocompact subgroup. If $\tilde\Gamma$ is a finite index supergroup of $\Gamma$, and if $\tilde\Gamma$ has no finite normal subgroups, then the given isometric action of $\Gamma$ extends to a faithful isometric action of $\tilde\Gamma$.
To prove it, since $\tilde\Gamma$ is quasi-isometric to $\Gamma$ which is quasi-isometric to $H^n$, applying the Sullivan-Tukia quasi-isometric rigidity theorem for $H^n$ it follows that there is a cocompact discrete action of $\tilde\Gamma$ on $H^n$ with finite kernel, but we have hypothesized the kernel away so this action is faithful. We now have two discrete cocompact actions of $\Gamma$: the given one, and the restriction of the one on $\tilde\Gamma$ that we got by applying Sullivan-Tukia. We can therefore apply Mostow rigidity and conjugate the $\tilde\Gamma$ action by an isometry of $H^n$, so that its restriction to $\Gamma$ agrees with the given $\Gamma$ action, and we are done.
Basically the same argument will work whenever the symmetric space satisfies Mostow rigidity and some reasonably strong form of quasi-isometric rigidity, which is true of just about every symmetric space associated to an irreducible semi-simple Lie group (except for $SL(2,R)$ of course).
And if it is OK for the action to have a finite kernel, then just drop the hypothesis that $\Gamma$ has no finite index normal subgroup, and the same argument still works.
ADDED AFTER FURTHER COMMENTS: See Misha's comment for an improvement on this argument that avoids QI-rigidity altogether, using solely Mostow rigidity.