The answer is NO for general non-selfadjoint subalgebras. With a subspace $X\subset \mathcal{B}(H)$, one can associate a subalgebra
$$S_X= \langle \begin{pmatrix} \alpha & x \cr 0 & \alpha \end{pmatrix} : \alpha \in \mathbb{C}I_H,\ x \in X \rangle \subset{\cal B}(H\oplus H).$$
Now choose a state $\omega$ which eliminates ${\cal K}(H)$ and let $X = \langle x \in B(H) : 2 x_{11} = \omega(x) \rangle$. Then, $\overline{S_X}=S_{\mathcal{B}(H)}$. However, since $x\in X\cap B_r$ implies $|x_{11}|\le r/2$, one has $\overline{S_X\cap B_r} \neq \overline{S_X}\cap B_r$.

In passing, I'll explain why Kaplansky's density theorem holds for $\mathrm{C}^\ast$-algebras. If $S \subset \mathcal{B}(H)$ is a subspace, then the inclusion extends to a weak$^\ast$-ultraweak continuous contraction $\pi\colon S^{\ast\ast}\to\mathcal{B}(H)$. By continuity and the Goldstine's theorem, one has $\overline{S\cap B_r}=\pi(S^{\ast\ast}\cap B_r)$. But if one knows $S^{\ast\ast}$ is actually a $\mathrm{C}^\ast$-algebra and $\pi$ is a $\ast$-homomorphism, then one gets $\pi(S^{\ast\ast}) \cong S^{\ast\ast}/\ker\pi$ isometrically, which implies $\pi(S^{\ast\ast}\cap B_r)=\pi(S^{\ast\ast})\cap B_r = \overline{S}\cap B_r$. This proof is probably circular, because Kaplansky's density theorem would be needed to justify some results quoted above.