Suppose that $u\in S_{U}$. Then I claim that $\inf\{\|u-v\|_{L^{2}(U)}:v\in S_{B}\}$ is bounded below by the standard deviation of the spherical symmetrization $u^{\sharp}$ of $u$.

For this post, the $L^{2}(U)$ norm will be with respect to the normalized area probability measure on $U$. Let $\mu$ be the Haar probability measure on the group of all $3\times 3$ orthogonal matrices. Let $\nu$ be the normalized area probability measure on $S^{2}$.

Define the spherical symmetrization $w^{\sharp}$ of a function $w$ by letting $$w^{\sharp}(x)=\int_{A\in O(3)}(w\circ A)(x)d\mu(A).$$ Observe that $$w^{\sharp}(x)=\int_{y\in S^{2}}w(\|x\|\cdot y)d\nu(y).$$

Observe that if $w$ is harmonic, then $w^{\sharp}$ is also harmonic, and there are constants $\alpha,\beta$ such that $w^{\sharp}(x)=\frac{\alpha}{\|x\|}+\beta$ (this fact generalizes to all dimensions $n\geq 2$).

By Jensen's inequality, if $r\in[0,1)$ and $x\in S^{2}$, then
$$(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}=(\int_{y\in S^{2}}f(ry)-g(ry)d\nu(y))^{2}\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$
Therefore, by integrating, we obtain
$$\int_{x\in S^{2}}(f^{\sharp}(rx)-g^{\sharp}(rx))^{2}dx\leq\int_{y\in S^{2}}(f(ry)-g(ry))^{2}d\nu(y).$$

Therefore, if $f,g:U\rightarrow\mathbb{R}$ are continuous, then
$\|f^{\sharp}-g^{\sharp}\|_{L^{2}(U)}\leq\|f-g\|_{L^{2}(U)}$

Suppose $u\in S_{U},v\in S_{B}$. Since $v$ is harmonic on $B$, the function $v$ satisfies the mean-value property, so the function $v^{\sharp}$ is constant.

Therefore, $$\text{Var}(u^{\sharp})\leq\|u^{\sharp}-v^{\sharp}\|_{L_{2}(U)}^{2}\leq\|u-v\|_{L^{2}(U)}^{2}.$$

There are plenty of functions $u$ that are harmonic on $U$ but where
$u^{\sharp}$ is non-constant on $U$ (such as the Newtonian potential), and for each such function, we have $$\text{Var}(u^{\sharp})>0.$$ This proof generalizes to any dimension $n\geq 2$ where the balls $B,B\setminus U$ have any radii but are still centered at $0$.