**No**, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be naturally identified with the space of all Pettis integrable functions from the domain of $\mu$ to $X$ (See Proposition 3.13 in Ryan's *Introduction to tensor products of Banach sapces*). However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional.

It seems to me that Pettis integrable functions form a closed subspace of the space of Dunford integrable functions, hence you may extend the above result as in the case where $X$ is infinite-dimensional, you have an incomplete, closed subspace of a normed space, so the space itself cannot be complete.