Assuming $B(x,r)$ denotes the open ball about $x$ of radius $r$, this map is lower semicontinuous but need not be upper semicontinuous.
Note that if $d(x,y) < r$ then $d(x_n,y) < r$ for sufficiently large $n$, so $y \in B(x_n, r)$. This means $\liminf_{n \to \infty} 1_{B(x_n, r)}(y) \ge 1_{B(x, r)}(y)$. Now we can use Fatou's lemma to see $\liminf_{n \to \infty} \int 1_{B(x_n, r)}\,d\mu \ge \int \liminf_{n \to \infty} 1_{B(x_n, r)}\,d\mu \ge \int 1_{B(x,r)}\,d\mu.$ That is, $\liminf_{n \to \infty} f(x_n) \ge f(x)$.
To see it need not be upper semicontinuous, take $X = \mathbb{R}$, $\mu = \delta_0$, $x_n = 1 - 1/n$, $x=1$, $r=1$. Then $\mu(B(x_n, r)) = 1$ for all $n$ but $\mu(B(x,r)) = 0$.
In fact, $f$ is upper semicontinuous (hence continuous) at $x$ if and only if $\mu(\partial B(x,r)) = 0$.
If $B(x,r)$ was meant to be the closed ball, then $f$ is indeed upper semicontinuous. The proof is essentially the same as above.