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).
It's a great question! Disappointingly, I think the answer to (2) is No :
The only restriction on a `good' division into "symmetric" vs. "symplectic" conjugacy classes that I can see is that it should be intrinsic, depending only on $G$ and the class up to isomorphism. (You don't just want to split the self-dual classes randomly, right?) This means that the division must be preserved by all outer automorphisms of $G$, and this is what I'll use to construct a counterexample. Let me know if I got this wrong.
The group
My $G$ is $C_{11}\rtimes (C_4\times C_2\times C_2)$, with $C_2\times C_2\times C_2$ acting trivially on $C_{11}=\langle x\rangle$, and the generator of $C_4$ acting by $x\mapsto x^{-1}$. In Magma, this is G:=SmallGroup(176,35), and it has a huge group of outer automorphisms $C_5\times((C_2\times C_2\times C_2)\rtimes S_4)$, Magma's OuterFPGroup(AutomorphismGroup(G)). The reason for $C_5$ is that $x$ is only conjugate to $x,x^{-1}$ in $C_{11}\triangleleft G$, but there there are 5 pairs of possible generators like that in $C_{11}$, indistinguishable from each other; the other factor of $Out\ G$ is $Aut(C_2\times C_2\times C_4)$, all of these guys commute with the action.
The representations
The group has 28 orthogonal, 20 symplectic and 8 non-self-dual representations, according to Magma.
The conjugacy classes
There are 1+7+8+5+35=56 conjugacy classes, of elements of order 1,2,4,11,22 respectively. The elements of order 4 are (clearly) not conjugate to their inverses, so these 8 classes account for the 8 non-self-dual representations. We are interested in splitting the other 48 classes into two groups, 28 'orthogonal' and 20 'symplectic'.
The catch
The problem is that the way $Out\ G$ acts on the 35 classes of elements of order 22, it has two orbits according to Magma - one with 30 classes and one with 5. (I think I can see that these numbers must be multiples of 5 without Magma's help, but I don't see the full splitting at the moment; I can insert the Magma code if you guys want it.) Anyway, if I am correct, these 30 classes are indistinguishable from one another, so they must all be either 'orthogonal' or 'symplectic'. So a canonical splitting into 28 and 20 cannot exist.
Edit: However, as Jack Schmidt points out (see comment below), it is possible to predict the number of symplectic representations for this group!
Best Answer
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.