$U$ need not be unitary. Let $U_n$ be the map on $\ell^2$ that sends $e_j\mapsto e_{j+1}$ for $j=1,\ldots, n$, $e_{n+1}\mapsto e_1$, and $U_n$ is the identity on $L(\{e_j: j>n+1\})$.
Then $U_n$ is unitary (it sends an ONB to an ONB again), but $U=S$ (the shift) is only an isometry because $e_1$ is not in the range.
The answer is a resounding no. In fact, the wot-closure of the set of unitaries is the whole unit ball.
The proof is a generalization of the idea, with matrices, that given any $z\in\overline{\mathbb D}$, there exists a unitary $$ \begin{bmatrix} z&*\\ *&*\end{bmatrix}.$$ One of many possible choices is $$\begin{bmatrix}z&(1-|z|^2)^{1/2}\\(1-|z|^2)^{1/2}&-z\end{bmatrix},$$that we will mimic below.
Concretely, let $x\in B(H)$ be a contraction, i.e. $\|x\|\leq 1$. Neighbourhoods in the wot topology are of the form $$N=\{y:\ |\langle (x-y)\xi_j,\eta_j\rangle|\leq1:\ \xi_1,\ldots,\xi_m,\eta_1,\ldots,\eta_m\in H\}.$$
Fix one such $N$ and let $$L=\operatorname{span}\{\xi_1,\ldots,\xi_m,\eta_1,\ldots,\eta_m%,x\xi_1,\ldots,x\xi_m,x\eta_1,\ldots,x\eta_m
\}.$$
Let $p$ be the orthogonal projection onto $L$. Since $\dim L<\infty$, there exists a subspace $L'\subset L^\perp$ with $\dim L'=\dim L$. Let $q$ be the orthogonal projection onto $L'$, and $v$ a partial isometry $v:L'\to L$, i.e. $v^*v=q$, $vv^*=p$. Now based on the decomposition $H=L\oplus L'\oplus ( L\oplus L')^\perp$, let $$u=pxp + (p-(pxp)^*pxp)^{1/2}v+ v^*(p-(pxp)^*pxp)^{1/2} - v^*pxpv+I-p-q.
$$
In spirit,
$$u=\begin{bmatrix} pxp & (p-(pxp)^*pxp)^{1/2}v&0\\ v^*(p-(pxp)^*pxp)^{1/2} &- v^*pxpv&0\\ 0&0&I\end{bmatrix}.
$$
It is not hard to check that $u$ is a unitary. And
$$
\langle u\xi_j,\eta_j\rangle
=\langle up\xi_j,p\eta_j\rangle
=\langle pup\xi_j,\eta_j\rangle
=\langle pxp\xi_j,\eta_j\rangle
=\langle x\xi_j,\eta_j\rangle.
$$
So $u\in N$. As we can do this for any neighbourhood $N$ of $x$, we can construct a net $u_N$ of unitaries with $u_n\to x$ wot.
This argument provides a way to show that the wot and sot topologies do not agree (and are, in fact, very different). The surprising thing, though, is that if $\{u_j\}$ are unitaries and $u_j\xrightarrow{\ \rm wot\ }u$ with $u$ unitary, then $u_j\xrightarrow{\ \rm sot\ }u$.
Best Answer
If it doesn't converge strongly then it doesn't converge in norm (consider the contrapositive), so try to find a counterexample there. To show that $\{S^n\}$ does not converge strongly to $0$ consider the vector $e_1\in\ell^2$ defined by $e_1(n)=1$ if $n=1$ and $e_1(n)=0$ otherwise.