Assume that $\det\Sigma\ne0$. Then the random matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the [Moore--Penrose inverse][1] of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence 
$$X^+X=X^\top(XX^\top)^{-1}X.$$
Letting 
$$Z:=S^{-1/2}X,$$
where $S:=I_m\otimes\Sigma$ (the Kronecker product), we see that the rows $z_i$ of the matrix $Z$ are iid $N(0,I_n)$, and 
$$X^+X=X^\top(XX^\top)^{-1}=Z^\top(ZZ^\top)^{-1}Z=Z^+Z.$$
So, the problem reduces to the case when $\Sigma=I_n$, which you said is easy. 

  [1]: https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Rank_decomposition