Think of $M$ first as a linear operator acting on $V = \mathbb{Q}^n$. Pass to a splitting field $K$ and consider the induced action on $V \otimes K$. This splits up into a direct sum of generalized eigenspaces of $M$, which are also then permuted by the Galois action of $G = \text{Gal}(K/\mathbb{Q})$ into orbits. By Galois descent we can find $G$-invariant bases of these sums over each Galois orbit of generalized eigenspaces. With respect to these bases, $M$ acts on each orbit by a Galois-irreducible rational matrix. If $M$ has no nontrivial Jordan blocks then you can replace "generalized eigenspaces" with "eigenspaces" and I think with a little more fiddling you get your second statement as well.
So that's over $\mathbb{Q}$. I'm less sure what to do over $\mathbb{Z}$. Presumably you should look at the induced action on $\mathbb{Z}^n \otimes \mathcal{O}_K$.
Edit: Again over $\mathbb{Q}$, you can avoid appealing to Galois descent as follows: the structure theorem for f.g. modules over a PID implies that $V$ above, as a $\mathbb{Q}[x]$-module (where $x$ acts by $M$), breaks up as a direct sum of modules of the form $\mathbb{Q}[x]/f(x)^m$ where $f$ is irreducible. Multiplication by $x$ always acts by a Galois-irreducible matrix on such submodules. Trivial Jordan blocks means we can take $m = 1$.