Hello again. Yes, it's true.

The more general statement you want is, let $X$ be projective with a
$K$-equivariant ample line bundle ${\mathcal O}(1)$. For each $n$, let $\mu_n$
be a measure on ${\mathfrak t}^*_+$,
$$ \mu_n := \sum_{\lambda \in {\mathfrak t}^*_+}
\frac{\dim Hom_K(V_{n\lambda}, \Gamma(X;{\mathcal O}(n))}{n^{\dim X}}
\delta_{\lambda}. $$
(Note that $V_{n\lambda}$ only means anything if $n\lambda$ is integral.)
Then $\lim_{n\to \infty} \mu_n$ is the nonabelian DH measure for
$K$ acting on $X$.

One approach to proving this to degenerate $X$
to $X' := (X//N \times G//N) // T$,
where $G = K^{\mathbb C}$ and $N$ is a maximal unipotent group.
Then the nonabelian DH measure of $X$ is the nonabelian DH measure of $X'$
is the {\em abelian} DH measure of $X//N$. (Here $G//N$ goes by the name
"Gel$'$fand variety", $X//N$ by "imploded cross-section", and
$X'$ by "Vinberg asymptotic cone".) At that point the theorem has to
be checked once and for all for $G//N$.

Really, the point is that on both the symplectic and quantum sides one
can obtain the nonabelian measure from the abelian, which sounds
surprising but is just another way of saying that characters
characterize representations. On the quantum (resp. symplectic) side, take the
weight multiplicity function (resp. abelian DH measure) and apply
differencing operators (resp. directional derivatives) in the
directions of positive roots.