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 $L^0$.

Now I claim that

> The distribution of the $L^0$ version of $X$ can be recovered from the joint distribution of $(X_t, t \ge 0)$.

Note that this will immediately imply a positive answer to your question.

Here is a sketch of a proof of my claim. Consider the functions $L^0(\mathbb{R}_+, E) \to \mathbb{R}$ of the following form:

$$F_{f,T}: x \mapsto \intop_0^T f(x_t) dt,$$

where $f: E \to [0,1]$ is bounded and Borel, and $T \ge 0$. These functions generate the whole Borel $\sigma$-algebra on $L^0$, because the level sets $\{F_{f,T} < \mathrm{const}\}$ for continuous $f: E \to [0,1]$ generate the topology.

Now it remains to pin down the joint distribution of $F_{f,T}(X)$. This is done by observing that the moments can be calculated using the joint distribution of $(X_t)$ by Fubini's theorem:

$$\mathsf{E} \intop_0^T f_1(X_t) dt \dots \intop_0^T f_n(X_t) dt = \intop_{[0,T]^n} \mathsf{E} f_1(X_{t_1}) \dots f_n(X_{t_n}) dt_1 \dots dt_n$$