The answer is: yes iff $X$ contains a finite dimensional invariant subspace of arbitrary large dimension (as you see, it has really nothing to do with the Banach structure of $X$ only with its quality as a $\Gamma$-representation).
Equivalently, the answer is: yes iff $X_0$ (defined below) is of infinite dimension.

First example (which you will not like): take a trivial representation of $\Gamma$ on $X$ and any probability measure supported on $B_X$ (but not on a finite dimensional subspace).

You want to disregard the first example as non-ergodic.

Second example (very explicit): Take $\Gamma= \mathbb Z$ and let it act by an irrational rotation (by $\sqrt{2}$, as I said I'll be explicit) on $S^1$. Consider $X=L^2(S^1)$. In $B_X$ take the subset $Y$ consisting of all characteristic functions of half circles in $S^1$ (normalized). $Y$ is an $S^1$ equivariant copy of $S^1$ in $B_X$ and we can endow it with the Haar measure of $S^1$. This measure satisfies everything you wished for.

The second example is typical in the sense that the $\Gamma$-invariant measure is actually invariant under a compact group enveloping $\Gamma$.

In what follows I will discuss $\Gamma$-invariant measures on the Banach space $X$ (not necessarily on the unit ball). I will assume that $X$ is separable, which probably is enough for you (e.g if $\Gamma$ is countable, or more generally, if $\Gamma$ is locally compact second countable and it acts strongly continuously on $X$). The answer above will follow easily from this discussion.

If $X$ has no finite dimensional invariant subspaces (other than $\{0\}$) then the only invariant probability measure is the Dirac measure at $0$
(for $X$ Hilbert this is Lemma 5.6 in "Amenable Invariant Random Subgroups by" Bader*-Duchesne-Lecureux).
More generally, let $X_0$ be the closure in $X$ of the vector space generated by all invariant finite dimensional subspaces. Then every invariant probability measure $\mu$ on $X$ is supported on $X_0$ and it is given as follows:

Fix a probability measure $\nu$ on the orbit space $X_0/G$, where $G$ is the Bohr compactification of $\Gamma$ (recall that the $\Gamma$ action on $X_0$ extends to $G$) and set
$$\mu=\int_{X_0/G} (\text{The normalized Haar measure on the orbit})~ d\nu(\text{orbit}).$$
 
In particular, every ergodic $\Gamma$-measure is supported (on $X_0$ and) on a unique $G$-orbit, and it is the Haar measure on it.
Note that by ergodic decomposition it is enough to prove the last statement.

Claim: let $\Gamma$ be a group acting isometrically on a complete, separable metric space $Y$ preserving a fully supported, ergodic probability measure $\mu$. Then there is a $\Gamma$-equivariant homeomorphism $Y\simeq G/L$, where $G$ is the Bohr compactification of $\Gamma$ and $L<G$ a closed subgroup, and under this identification $\mu$ corresponds 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). Then the isometry group of $Y$ is compact by Arzela-Ascoli and the $\Gamma$ action extends to $G$. By ergodicity (and the fact that $Y/G$ is nice) we get that $\mu$ is supported on a unique orbit and everything else follows easily.

Now, for the statement we had regarding ergodic measures on $X$, take $Y$ to be the support of an ergodic measure on $X$, which I assume to be total. Observe that the action of $G$ on $Y$ extends to a linear isometric action on $X$. Call Peter-Weyl to see that $X=X_0$.

$*$ Hi.