No. $\rm HOD$ is a model of $\sf ZFC$, therefore it has sets of reals that it thinks are not Baire measurable.

Moreover, since $M[G]$ was obtained by the Levy collapse, $\mathrm{HOD}^{M[G]}=\mathrm{HOD}^M$. So unless $M$ itself was a Solovay model, there's little to no chance that the argument will go through.

In particular, if you start with a nice ground model like $L$, then you can't get any "deeper" and $\rm HOD(HOD)$ is a moot construction.

There's not enough information in the question. If $M\models V=L$, then in $M$ we have that $\mathrm{HOD}^M=L^M=M$, and therefore $\mathrm{HOD}^{\mathrm{HOD}^M}=L^M=M$ as well.

In particular we have that in both instances sets of real definable from ordinals and reals (in $M$, which is the same as $\mathrm{HOD}^M$ in this case) are the same sets of reals in $L$, and so there are sets without the Baire property.