It was asked in the comments that I provide some details. I prove slightly more: if $\mu_n$ denotes uniform probability on the **sphere** of radius $n$ and if $\rho:F_2 \to \mathrm{SU}_2$ is a homomorphism with dense image, then for any continuous function $f$ on $\mathrm{SU}_2$,
$$ \lim_n \int f(\rho(a)) d\mu_n(a) =\int f.$$

For the proof, consider the unitary representation $\pi$ on $L_2(\mathrm{SU}_2)$, $\pi(a) f(x) = f( \rho(a^{-1}) x)$. The assumption that $\rho$ has dense image gives that the $\pi(F_2)$-invariant functions are the constant functions. By strict convexity of $L_2$, this can be translated to: the eigenvectors of $A:=\pi(\mu_1) = \int \pi(a) d\mu_1(a)$ with eigenvalue $1$ are the constant functions. By similar argument, $A$ does not have $-1$ has an eigenvalue.

Now a classical computation shows that $\mu_1 \ast \mu_n = \frac{3}{4} \mu_{n+1} + \frac 1 4 \mu_{n-1}$ for every $n \geq 1$, and therefore we can write $\mu_n = P_n(\mu_1)$ where $P_n$ is the polynomials defined by $P_0=1$, $P_1=X$ and $X P_n = \frac{3}{4} P_{n+1} + \frac 1 4 P_{n-1}$. Solving the recurrence we see that $P_n(1)=1$ and $\lim_n P_n(x)=0$ for every $x \in (-1,1)$. By the spectral theorem, we obtain that $\lim_n \pi(\mu_n)f = \int f$ in $L_2(\mathrm{SU}_2)$ (and in particular in probability) for every $f \in L_2(\mathrm{SU}_2)$.

Finally, if $f$ is continuous, the family of functions $\pi(\mu_n)f$ is uniformly equicontinuous, so convergence in probability implies uniform (and in particular pointwise at the identity) convergence, QED.

As explained by Lucas Kaufmann, in some situations it is known that $A$ has spectral gap, so the convergence of $\pi(\mu_n)$ to the orthogonal projection on the constants happens in operator norm, and not only pointwise.