You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$, with characters $\chi_X$ and $\chi_Y$, then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \otimes \chi_Y$, where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$. The proof of this is exactly as in the finite case.
Now, let $W=1 \oplus V \oplus \overline{V}$. Your goal is to show that, for any nonzero representation $Y$, $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else.
I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$. The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$, of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$, we have $f=1+2 \sum \cos \theta_i$. In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have
$$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$
where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$.
And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$. So the contribution to our integral near the identity can be approximated by
$$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$
where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$. You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$, I'll leave that to you as well.
For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial.
Now, look at the contribution from the rest of $G$. For any point in the unitary group $U(d)$, other than the identity, we have $|f| < 2n+1$. (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$.
So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar.
Frobenius reciprocity for Brauer characters is a little more complicated (a lot more complicated if you don't have complete tables). You need the projective characters to compute the multiplicities. I don't use anything special about the character being induced, though sometimes you can leverage that information (especially if you don't have complete tables).
In GAP, this is easily done:
g:=CharacterTable("ON");
# Let g be the O'Nan simple group
chi:=Sum([1..10],i->Random(Irr(g mod 2)));;
# chi is a random reducible Brauer character
ipr:=Irr(g)*DecompositionMatrix(g mod 2);;
# ipr is the list of projective Brauer characters
vec:=MatScalarProducts(ipr,[InducedClassFunction(chi,g)])[1];
# vec is the multiplicity of each Brauer character in chi
chi = vec * Irr( g mod 2 );
# should be true: we have decomposed it correctly.
To get the decomposition matrix, you must not only have the Brauer table but also the ordinary table. You don't need much from the subgroup H, just a character to induce and the element fusion from H into G.
This is theorem 2.13 on page 25 of Navarro's textbook on Characters and Blocks of Finite Groups.
By projective, I mean in the module theory sense, an indecomposable direct summand of the regular representation, usually these are denoted Φi corresponding to a Brauer character φi. I don't mean representation into projective groups, like PGL.
Best Answer
If you want a representation $V$ to be self-dual up to a character, then either $S^2V$ or $\Lambda^2V$ (considered as representations of $G$) should have a 1-dimensional summand (corresponding to the isomorphism $V \to V^*\otimes\chi$). But as it was mentioned by Jim there are a lot of representations $V$ for which both $S^2V$ and $\Lambda^2V$ are irreducible. For example, this is the case for $G = \operatorname{SL}(n)$ and $V$ the standard representation — the example suggested by Bruce.