The answer is clearly "yes" if you don't restrict to ergodic measures: take even a trivial action of $\Gamma$ and whatever probability measure on the sphere.
If you restrict to ergodic measures, then the answer is "no". Let me assume here that the Banach space is separable (a reduction to this case is automatic if $\Gamma$ is countable).
Claim: let $\Gamma$ be a group acting isometrically on a metric space $Y$ preserving an ergodic probability measure $\mu$. Assume $Y$ is separable and complete and $\mu$ is fully supported. Then: $Y$ is compact, the isometry group of $Y$ is compact, the closure of $\Gamma$ in it, $K$, preserves $\mu$ and acts transitively on $Y$, thus $Y\simeq K/L$ for some closed subgroup $L<K$ and under this identification $\mu$ correspond to the Haar measure.
The essential part of the claim is the compactness of $Y$, which follows by completeness from total-boundedness, which is easy (generalize the standard claim that $Y$ is bounded, which is proven by taking a compact set of more than half the weight). Everything else follows: the compactness of the isometry group by Arzela-Ascoli, the transitivity of $K$ by its compactness and the ergodicity of $\mu$.
Now, for the asked question, take $Y$ to be the support of $\mu$ in $X$, which I assume to be total. Observe that the action of $K$ extends to an isometric, hence linear, action on $X$. Call Peter-Weyl.