Let $G$ be a finite group. Then the irreducible complex representations of $G$ come in three sorts: real, complex and symplectic=quaternionic. The type of an irreducible character $\chi$ can be read of from the Frobenius-Schur indicator 
$$ s_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi( g^2 ) \in \{ 1,0,-1 \}. $$
Now the following seems to be true (and has been used by Noah Snyder in his interesting answer to  [another question][squareroots]) but I can't see why:
Suppose all characters of $G$ have real values. (Equivalently, every element of $G$ is conjugate to its inverse.) Then it seems that the Frobenius-Schur indicator defines a grading of the irreducible representations, and thus of the character ring. This means that if $\chi$ and $\psi$ are irreducible characters with $s_2 (\chi) = s_2(\psi)$, then all the irreducible constitutents of $\chi\psi$ have indicator $1$, and if $s_2(\chi) = -s_2(\psi)$, then all constituents of $\chi\psi$ have indicator $-1$. Why is this actually true?  Of course, for example in the first case, $\chi\psi$ is afforded by a real representation, so the symplectic representations must occur with even multiplicity. But why can they not occur at all?  

Moreover, I would like to know if this generalizes to other fields than $\mathbb{R}$, using elements of a Brauer group instead of the Frobenius-Schur indicator.  
Edit: For some hours I thought I had a counterexample to this generalization, but this was wrong. It seems to be difficult to find a group where all characters have values in a field $\mathbb{F}$ and such that the endomorphism ring of some simple $\mathbb{F}G$-module is a division ring other than $\mathbb{F}$ or the quaternions over $\mathbb{F}$.

[squareroots]: http://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory