The proof of Burnside-Brauer adapts easily to give the following result for Lie groups:
Let $G$ be a compact semi-simple Lie group and let $R$ be a faithful irrep of $G$. Let $U$ be any irrep. Then $U$ occurs in $R^{\otimes M}$ for some $M$.
Some preliminary comments:
Since $G$ is compact, the Lie algebra of $G$ is the direct sum of a semi-simple Lie algebra and an abelian Lie algebra; so the semi-simple hypothesis just says that there is no abelian summand.
Since $G$ is compact, we may assume that $R$ is a unitary representation, by the standard averaging argument. Since the representation is faithful, we may thus view $G$ as a subgroup of a unitary group. Let $n= \dim R$ and consider $G$ as a subgroup of $U(n)$ from now on.
Let $Z$ be the center of $G$. By Schur's lemma, $Z$ acts on $U$ and $R$ by scalars; let $\psi_R$ and $\psi_U$ be the characters by which $Z$ acts on $R$ and $U$. Since $G$ is semi-simple, $Z$ is discrete. Now, $Z$ lies in the center of $U(n)$, which is $S^1$, so $Z$ must be a cyclic group; say $Z \cong \mathbb{Z}/m$. Let $\psi_U = \psi_R^h$. What we will actually be showing is that $\mathrm{Hom}(U, R^{\otimes (h+mN)})$ is nonzero for all sufficiently large $N$.
Proof sketch: Let $\mu_G$ be Haar measure, normalized so that $\int_G \mu_G=1$. Then $$\dim \mathrm{Hom}(U, R^{\otimes (h+mN)}) = \int_G \chi_R^{h+mN} \overline{\chi_U} \mu_G$$ where $\chi_R$ and $\chi_U$ are the characters of $U$ and $R$.
Since $\chi_R$ is the character of a subgroup of $U(n)$, we have $|\chi_R(g)| \leq n$, with equality only when $g$ is of the form $e^{i \theta} \mathrm{Id}$. Such $g$'s are precisely the elements of $Z$. So, for large $N$, the integral is dominated by the contributions from small neighborhoods of the points of $Z$. For $z \in Z$, we have $\chi_R(z)^m = n^m$ and $\chi_r(z)^h \overline{\chi_U}(z) = n^{h+1}$. If you work out the asymptotics of the integrals, using the [method of steepest descent][1], you get the contribution from each $z \in Z$ is of the form $$c n^{mN} N^{-\dim G/2} (1+O(N^{-1/2}))$$ for some $c>0$.
In particular, for $N$ large, all these contributions are positive and $\dim \mathrm{Hom}(U, R^{\otimes(h+mN)})$ is nonzero. $\square$
Commment 1 I'd love to know whom to credit for this trick. I came up with it in this [blog thread][2] and used it again [here][3] but I'm sure it's not original to me.
Comment 2 If $G$ does have an abelian factor, life is harder. The obvious counterexample is that, if $G$ is $S^1$, the irrep $R$ is $\theta \mapsto e^{i \theta}$ and $U$ is $\theta \mapsto e^{-i\theta}$, then $U$ is not a summand of $R^{\otimes n}$ for any $n \geq 0$. More subtly, let $G = U(3)$, let $V$ be the standard three dimensional irrep and let $R=\mathrm{Sym}^2 V$ and $U = \bigwedge^2 V$. Then $U$ is not a summand of $R$, and the matrix $e^{i \theta} \mathrm{Id}$ acts on $R^{\otimes n}$ by $2n \theta$ and acts on $U$ by $2 \theta$, so $U$ is not a summand of $R^{\otimes n}$ for $n>1$.
Comment 3 If $G$ is a complex semi-simple group, the same result holds for $G$ by the standard nonsense relating complex and compact representations.
Comment 4 If $G$ is a real semi-simple group, life is more confusing. For example, let $G = SL_3(\mathbb{R})$, let $V$ be the standard three dimensional rep of $G$, and let $R = \mathrm{Sym}^3 V$. Then $G \to GL(R)$ is injective, but the corresponding complex representation $SL_3(\mathbb{C}) \to GL(R \otimes \mathbb{C})$ is not. One should be able to formulate a statement here with a bit more care, but I'm not going to do it.
Comment 5 I have no idea what the answer is if $G$ is not semi-simple. [1]: http://en.wikipedia.org/wiki/Method_of_steepest_descent [2]: http://sbseminar.wordpress.com/2010/11/25/passage-from-compact-lie-groups-to-complex-reductive-groups/ [3]: Does every irreducible representation of a compact group occur in tensor products of a faithful representation and its dual?