Here are some things you probably know. For a representation $W$ of $G$, let $\text{Inv}(W)$ denote the subspace of $G$-invariants. For an irreducible representation $V$ with character $\chi$, the F-S indicator $s_2(\chi)$ naturally appears in the formulas
$$\dim \text{Inv}(S^2(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^2 + \chi(g^2)}{2}$$
and
$$\dim \text{Inv}(\Lambda^2(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^2 - \chi(g^2)}{2}.$$
More precisely the F-S indicator is their difference, while their sum is $1$ if $V$ is self-dual and $0$ otherwise. The corresponding formulas involving $s_3(\chi)$ are
$$\dim \text{Inv}(S^3(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^3 + 3 \chi(g^2) \chi(g) + 2 \chi(g^3)}{6}$$
and
$$\dim \text{Inv}(\Lambda^3(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^3 - 3 \chi(g^2) \chi(g) + 2 \chi(g^3)}{6}.$$
Here the F-S indicator $s_3(\chi)$ naturally appears in the sum, not the difference, of these two dimensions. $T^3(V)$ decomposes into three pieces, and the third piece is (Edit, 9/26/20: two copies of) the Schur functor $S^{(2,1)}(V)$, which therefore satisfies
$$\dim \text{Inv}(S^{(2,1)}(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{ \chi(g)^3 - \chi(g^3)}{3}.$$
So $s_3(\chi)$ constrains the dimensions of these spaces in some more mysterious way than $s_2(\chi)$ does. The sum
$$\dim \text{Inv}(T^3(V)) = \frac{1}{|G|} \sum_{g \in G} \chi(g)^3$$
tell us whether $V$ admits a "self-triality," and this dimension is an upper bound on $s_3(\chi)$. If $V$ is self-dual, this is equivalent to asking whether there is an equivariant bilinear map $V \times V \to V$, which might be of interest to somebody. If this dimension is nonzero then $s_3(\chi)$ gives us information about how a triality behaves under permutation.
The situation for higher values of $3$ is worse in the sense that the bulk of the corresponding formulas are not completely in terms of F-S indicators but in terms of inner products of F-S indicators and their interpretation will only get more confusing. Already I don't know of many applications of triality (in fact I know exactly one: http://math.ucr.edu/home/baez/octonions/node7.html).
As Zoltan suspected in his answer, the statement in my question is not true. I have now found counterexamples: Let $q$ be an odd prime power. Then $SL(2,q)$ contains a group isomorphic to the quaternion group $Q_8$, which yields a (semiregular) action of $Q_8$ on $V=(\mathbb{F}_q)^2$.
Let $G = V \rtimes Q_8$ be the semidirect product. All faithful irreps of $G$ have degree $8$
(they are induced from nontrivial linear characters of $V$), and are of real type.
The other irreps have $V$ in its kernel and come from irreps of $Q_8$. In particular, all irreps of $G$ are self-dual, and exactly one is of symplectic type, of degree $2$. The tensor product of a faithful irrep with itself contains the symplectic irrep as summand.
Note that $\mathbf{Z}(G)=1$, so there's no nontrivial grading of the irreps of $G$.
Taking $q=3$ in the above yields a counterexample of order $72$, and I have to admit that I used GAP to find it. For the record, lets note that there are also four groups of order $64$, where all characters are real valued, but the FS-indicator doesn't define a grading of the irreps.
These are the groups with identifiers [64, 218], [ 64, 224 ], [ 64, 243 ], [ 64, 245 ] in the SmallGroups library of GAP.
Best Answer
Here's one characterization that I learned from Serre (see Definition 7.1.1 in his Topics in Galois Theory (p.65)): an element $g$ of a finite group $G$ satisfies $\chi(g) \in {\bf Q}$ for all characters $\chi$ iff $g$ is conjugate in $G$ to $g^m$ for all $m$ relatively prime to the exponent $e(g)$. [If $m$ is not coprime to $e(g)$ then $e(g^m) \lt e(g)$ so $g^m$ cannot possibly be conjugate to $g$.] It is enough to check this for all $m$ relatively prime to $\left| G \right|$. In particular, all character values are rational iff every group element is conjugate to its $m$-th power for all $m$ coprime to $\left| G \right|$.