Yes.

I assume that $E$ is a Polish space, or a standard Borel space that can be turned into a Polish space by some auxillary topology.

Basically, you can view your measurable stochastic process as a random element in the space $L^0(\mathbb{R}_+, E)$ of equivalence classes of $E$-valued measurable functions. $L^0(\mathbb{R}_+, E)$ comes equipped with the Polish topology of convergence in measure on $[0,n]$ for all $n \in \mathbb{N}$ and its corresponding standard Borel $\sigma$-algebra. While doing that you lose some information (e.g. the value for some specific $t$), but you will be able to calculate your Lebesgue measures as measurable functions on $L^0$.

Note that this random element in $L^0(\mathbb{R}_+, E)$ doesn't change if you replace the process by a version of itself. Namely, if $X$ and $X^\prime$ are versions of each other then for every measurable set $A \subset E$ the sets $\{t \mid X_t \in A\}$ and $\{t \mid X^\prime_t \in A\}$ will almost surely differ on a set of Lebesgue measure $0$, by Fubini's theorem:
$$\mathsf{E} \intop | \mathsf{1}\{X_t \in A\} - \mathsf{1}\{X^\prime_t \in A\} | dt = \intop \mathsf{E} | \mathsf{1}\{X_t \in A\} - \mathsf{1}\{X^\prime_t \in A\} | dt = 0$$

Now your stationarity assumption means that there is a measure-preserving transformation $T_s: \Omega \to \Omega$, such that $(X_{t+s}, t \ge 0)$ and $(X_t \circ T_s, t \ge 0)$ are versions of each other, so the $L^0$-valued random variables corresponding to them will be the same.