To summarize the situation given in the other answers (no real new content here) it is classical theory going back to Young that the complex irreducible representations of $S_n$ can be defined over $\mathbb Q$ (i.e., written with $\mathbb Q$-coefficients, or written as $\mathbb C\otimes_{\mathbb Q}V$ with $V$ a $\mathbb QS_n$-irreducible module) and references were given; i.e., the $\mathbb Q$-irreducibles are absolutely irreducible. This can be done via Young symmetrizers and anti-symmetrizers, polytabloids, or a number of other approaches and I have nothing to add to the discussion.
What this means concretely is that $\mathbb QS_n\cong \prod_{i=1}^{p_n}M_{d_i}(\mathbb Q)$ where $p_n$ is the number of partitions of $n$ and $d_i$ is the dimension of the $i^{th}$-irreducible representations (and of course all these $d_i$ are well known through tableaux combinatorics and involve counting semi-standard Young tableaux). One way to see this is to use that if $V$ is a finite dimensional $\mathbb QS_n$-module, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)\cong \mathbb C\otimes_{\mathbb Q}\mathrm{End}_{\mathbb QS_n}(V)$ by standard arguments and so by Schur's lemma, if $\mathbb C\otimes_{\mathbb Q}V$ is irreducible, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)$ one-dimensional over $\mathbb C$ and hence $\mathrm{End}_{\mathbb QS_n}(V)$ is one-dimensional over $\mathbb Q$ and so apply Wedderburn-Artin to $\mathbb QS_n$ to get the statement.
Now, let's just handle $\mathbb Q[S_n\times S_m]\cong \mathbb QS_n\otimes_{\mathbb Q}\mathbb QS_m$. Then by the above, we have that this tensor product is isomorphic to $$\prod_{i=1}^{p_n}\prod_{j=1}^{p_m}M_{d_i}(\mathbb Q)\otimes_{\mathbb Q} M_{c_j}(\mathbb Q)$$ where I introduced $c_j$ for the dimensions of the $S_m$-irreducibles over $\mathbb Q$. Obviously $$M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)\cong M_{d_i}(M_{c_j}(\mathbb Q))\cong M_{d_ic_j}(\mathbb Q)$$ and hence $M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)$ is simple with a unique simple module (up to isomorphism) which has dimension $d_ic_j$ (and this dimension characterizes the simple module). Consequently the tensor product of the unique simple modules of the two tensor factors is the unique simple module for this tensor product, e.g., by dimension consideration. Putting it all together, we get that the simple $\mathbb Q[S_n\times S_m]$-modules are the tensor products of the simple $\mathbb QS_n$-modules and $\mathbb QS_m$-modules. Now since $K\otimes_{\mathbb Q} M_r(\mathbb Q)\cong M_r(K)$ and $K\otimes_{\mathbb Q}\mathbb Q^n\cong K^n$, the situation doesn't change when we extend the scalars. We get the same number of irreducibles and they are obtained by extending the scalars from those of $\mathbb Q[S_n\times S_m]$. Of course the argument is the same for any finite number of factors.