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 isometrically embedded into the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces) and all simple functions can be regarded as elements of this tensor product. 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, so the space of Pettis integrable functions is complete only in this case.
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.
Addendum. Let me point out that if you want to do some functional analysis with the space of Pettis integrable functions, even though incomplete, it is barrelled.