There are counterexamples as soon as $g > 1$.

Let $n$ be the number of surjective homomorphisms $\pi_1(S) \to A_5$, up to $S_5$-conjugacy. (We can see that $n \geq 1$ using the fact that $A_5$ can be generated for two elements.)

Then there is a homomorphism $\pi_1(S) \to A_5^n$, where we take the product of all these homomorphisms. I claim this homomorphism is surjective.

Proof: Let $H$ inside $A_5^n$ be the image. Its projection onto each factor is surjective, and these surjective morphisms are distinct up to automorphisms of $A_5$. All such subgroups are all of $A_5^n$, by induction on $n$: Given that it is true for $A_5^{n-1}$, the projection of a subgroup $H$ onto the first $n-1$ factors must have image $A_5^{n-1}$, and the kernel is a normal subgroup of $A_5$, hence either $A_5$ or $1$. If $A_5$ then $H= A_5^n$, if $1$ then there is an $n$th homomorphism $H= A_5^{n-1} \to A_5^n$ which is not equal to any of the other ones up to an automorphism, which is false.

Now if we take $V$ an irreducible representation of $A_5$ which is stable under the action of $S_5$ (like the standard representation), then $V^{ \otimes n}$ is an irreducible representation of $A_5^n$ and thus by composition an irreducible representation of $\pi_1$. Any element of the mapping class group permutes the homomorphisms $\pi_1(S)$ (and maybe also acts on them by elements of $S_5$) and so preserves $V^{\otimes n}$, so $V^\otimes n$ is invariant under the mapping class group.

We could also run this argument with the Mathieu group $M_{11}$, which has no outer automorphisms at all, so any irreducible representation works.