I don't know the answer, but here are a few remarks :
(0) As has been pointed out before, "irreducible" and "indecomposable" are not the same for representations in positive characteristic. "Irreducible" is a stronger property. (For example, irreducible representations of commutative groups are always $1$-dimensional, whereas indecomposable representations don't have to be. No commutative group is going to give a counterexample.)
(1) The answer is "yes" if we look at representations over a field of characteristic prime to the order of $G$ (because representations theory of $G$ will be the same as in characteristic zero).
(2) If we want to construct a counterexample, then we should not take $G$ to be a $p$-group (as in the two attempts above). Because then $k$ will have to be of characteristic $p$, but then every finite-dimensional representation of $G$ over $k$ has a nonzero fixed vector.
(3) More generally, it is not possible to construct a counterexample in characteristic $p$ if $G$ has a normal $p$-Sylow $H$ (let $W$ be the subspace of vectors fixed by $H$, it is nonzero because $H$ is a $p$-group, it is stable by $G$ because $H$ is normal, hence it has to be the whole space because the representation is irreducible, but then we are reduced to the case of $G/H$, which has order prime to $p$, and see (1)).
That means, I'm afraid, no easy counterexamples with very small groups.