**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][1].


  [1]: http://www.ams.org/journals/proc/1992-114-03/S0002-9939-1992-1107271-2/S0002-9939-1992-1107271-2.pdf