Suppose $f_n:\mathbb{R} \to S^1$ are given by $f_n(t) = \mathrm e^{\mathrm i x_nt}$, where $x_n$ is some fixed real sequence. It is clear that whenever $x_n$ converges, $f_n$ must converge to $\mathrm e^{\mathrm i t \lim_{n\to \infty}x_n}$.
If $f_n$ converges pointwise to a continuous function, is it true that $x_n$ must converge?
Intuitively i think it is true, but i couldn't find a way to prove it.
Your conjecture is true, and you can prove it without assuming the limit is continuous, and relaxing pointwise convergence to almost sure convergence on some open interval.
Call the limit $f$.
Case 1: $x_n$ is unbounded. For any interval $I$ we have $\int_I f=\int_I \lim f_n=\lim \int_I f_n=0$ by dominated convergence. Therefore $f=0$. This contradicts the fact that we don't have convergence in probability to $f=0$.
Case 2: $x_n$ is bounded. Then it has a convergent subsequence, with limit $x^*$. If this wasn't the limit of the original sequence, there'd be another subsequence limit $x_+$. Then you'd have $f_n$ converge to both $f_*$ and $f_+$, which contradicts the uniqueness of almost sure limits.
PS: you cannot relax your assumptions to convergence on $\mathbb{Q}$. In fact, enumerate the rationals as $q_n$ and pick $x_n$ such that $f_n(q_i)=1, i\leq n$.