This fact doesn’t need $M$ to be hyperbolic. It just needs one general theorem about 3-manifold topology, namely the Scott core theorem.

Let $N\to M$ be the covering space corresponding to the subgroup $\Gamma$. Since $\pi_1(N)=\Gamma$ is finitely generated, the Scott core theorem guarantees a compact submanifold $N_0\subseteq N$ such that inclusion induces an isomorphism $\pi_1(N_0)\cong\pi_1(N)$; in particular, $\pi_1(N_0)$ is the surface group $\Gamma$.

It is a byproduct of the proof of the Scott core theorem that none of the boundary components of $N_0$ are spheres: the point is that any spherical boundary component bounds a ball in $N$, so can be filled in. Now, note that $N_0$ is not closed (since it is a compact submanifold of the non-compact manifold $N$), so $\partial N_0$ is non-empty. A component $S\subseteq \partial N_0$ is then embedded in $N$, hence immersed in $M$, as required.

Note that the argument does not build $S$ with $\Gamma=\pi_1(S)$. Dehn’s lemma does guarantee that $\pi_1(S)$ is a subgroup of $\Gamma$, and it must be a subgroup of finite index. With a little more work, one can argue that $N_0$ is an interval bundle, and hence deduce find an immersed surface representing $\Gamma$ itself, but this wasn’t required for the question as stated.

This fact doesn’t need $M$ to be hyperbolic. It just needs one general theorem about 3-manifold topology, namely the Scott core theorem.

Let $N\to M$ be the covering space corresponding to the subgroup $\Gamma$. Since $\pi_1(N)=\Gamma$ is finitely generated, the Scott core theorem guarantees a compact submanifold $N_0\subseteq N$ such that inclusion induces an isomorphism $\pi_1(N_0)\cong\pi_1(N)$; in particular, $\pi_1(N_0)$ is the surface group $\Gamma$.

It is a byproduct of the proof of the Scott core theorem that none of the boundary components of $N_0$ are spheres: the point is that any spherical boundary component bounds a ball in $N$, so can be filled in. Now, note that $N_0$ is not closed (since it is a compact submanifold of the non-compact manifold $N$), so $\partial N_0$ is non-empty. A component $S\subseteq \partial N_0$ is then embedded in $N$, hence immersed in $M$, as required.

Note that the argument does not build $S$ with $\Gamma=\pi_1(S)$. Dehn’s lemma does guarantee that $\pi_1(S)$ is a subgroup of $\Gamma$, and it must be a subgroup of finite index. With a little more work, one can argue that $N_0$ is an interval bundle, and hence find an immersed surface representing $\Gamma$ itself, but this wasn’t required for the question as stated.