Let $(\Omega, \mathcal{F})$ be a measurable space equipped with a strictly increasing filtration $(\mathcal{F}_{n})_{n \in \mathbb{N}}$. Let $P$ and $Q$ be probability measures on $(\Omega, \mathcal{F})$.
Suppose $P$ and $Q$ are equivalent in the sense that $P(A) = 0$ iff $Q(A) = 0$ for all $A \in \mathcal{F}$. Does the following result hold?
$\sup_{A \in \mathcal{F}}|P(A \mid \mathcal{F}_{n}) - Q(A \mid \mathcal{F}_{n})| \to 0$ as $n \to \infty$ almost surely with respect to $P$ and $Q$.
ADDENDUM: Suppose that $\mathcal{F}_n \nearrow \mathcal{F}$. By the Levy 0-1 Law, both $P(A \mid \mathcal{F}_n)$ and $Q(A \mid \mathcal{F}_n)$ tend to $\mathbf{1}_A$ almost surely. So for this case, all that needs to be shown is that $P(A \mid \mathcal{F}_n)$ and $Q(A \mid \mathcal{F}_n)$ tend to $\mathbf{1}_A$ uniformly in $A$.
ADDENDUM 2: Here is an attempt at a solution for the case $\mathcal{F}_n \nearrow \mathcal{F}$. I would appreciate feedback.
We show that $|P(A \mid \mathcal{F}_n) - Q(A \mid \mathcal{F}_n)|$ is bounded by a quantity that is independent of $A$ and that tends to $0$ as $n \to \infty$. Let $q = dQ/dP$, and abuse notation by writing $A = \mathbf{1}_A$.
It's easy to show using the basic properties of conditional expectations that $$E_Q(A \mid \mathcal{F}_n) = E_P(Aq \mid \mathcal{F}_n)(E_P(q \mid \mathcal{F}_n))^{-1}.$$
Hence, omitting the subscript from $E_P$ henceforth, we have $$P(A \mid \mathcal{F}_n) - Q(A \mid \mathcal{F}_n) = \frac{E(A \mid \mathcal{F}_n)E(q \mid \mathcal{F}_n) - E(Aq \mid \mathcal{F}_n)}{E(q \mid \mathcal{F}_n)}.$$
By the equivalence of $P$ and $Q$, $q >0$ a.s., and $E(q \mid \mathcal{F}_n) \to q$ a.s. by the Levy 0-1 Law. It suffices to show that the numerator on the RHS tends to $0$ uniformly in $A$.
Using the basic properties of conditional expectations again, we find after some manipulation that
$$|E(Aq \mid \mathcal{F}_n) - E(A \mid \mathcal{F}_n)E(q \mid \mathcal{F}_n)| = |E[A(q - E(q \mid \mathcal{F}_n)) \mid \mathcal{F}_n]| \\ \leq E[|A(q - E(q \mid \mathcal{F}_n))| \mid \mathcal{F}_n] \leq E[|q - E(q \mid \mathcal{F}_n)| \mid \mathcal{F}_n].$$
Now, $E[|q - E(q \mid \mathcal{F}_n)| \mid \mathcal{F}_n]$ is independent of $A$. Moreover, $|q - E(q \mid \mathcal{F}_n)| \to 0$ as $n \to \infty $ because $E(q \mid \mathcal{F}_n) \to q$. Finally, $$|q - E(q \mid \mathcal{F}_n)| \leq |q| + \sup_n(|E(q \mid \mathcal{F}_n)|) = : Z$$
and $EZ < \infty$ because $\{ E(q \mid \mathcal{F}_n) \}$ is uniformly integrable and hence bounded in $L^1$. The dominated convergence theorem for conditional expectation now yields
$$E[|q - E(q \mid \mathcal{F}_n)| \mid \mathcal{F}_n] \to 0,$$ and we are done.
If $\mathcal{F}=\lim_n\mathcal{F}_n$, we in fact have the following limit, $$ {\rm ess\,sup}_Z\lvert E_P[Z\mid\mathcal{F}_n]-E_Q[Z\mid\mathcal{F}_n]\rvert\to 0, $$ (almost surely), with the essential supremum taken over random variables $\lvert Z\rvert\le 1$ (you can restrict to $Z=1_A$ for sets $A\in\mathcal{F}$ if you prefer). Setting $q=dQ/dP$ and $q_n=E_P[q\mid\mathcal{F}_n]$ then $E_Q[Z\mid\mathcal{F}_n]=E[Zq\mid\mathcal{F}_n]/q_n$. \begin{align} \lvert E_P[Z\mid\mathcal{F}_n]-E_Q[Z\mid\mathcal{F}_n]\rvert &=\lvert E_P[Z(q_n-q)\mid\mathcal{F}_n]\rvert/q_n\\ &\le E_P[\lvert q_n-q\rvert\mid\mathcal{F}_n]/q_n \end{align} (almost surely). As the right hand side does not depend on $Z$, the inequality holds for the essential supremum. By martingale convergence, $q_n\to q > 0$ almost surely. So, it just remains to show that $E_P[\lvert q_n-q\rvert\mid\mathcal{F}_n]$ goes to zero almost surely. 'Dominated convergence' as stated in the question gives, for any $\epsilon > 0$, \begin{align} &E_P[1_{\{q_n-q > \epsilon\}}\mid\mathcal{F}_n]\to0\\ &E_Q[1_{\{q_n-q > \epsilon\}}\mid\mathcal{F}_n]\to0 \end{align} almost surely. So, $$ E_P[1_{\{q_n-q > \epsilon\}}(q_n-q)\mid\mathcal{F}_n]=q_nE_P[1_{\{q_n-q > \epsilon\}}\mid\mathcal{F}_n]-q_nE_Q[1_{\{q_n-q > \epsilon\}}\mid\mathcal{F}_n] $$ tends to zero almost surely and, \begin{align} E_P[\lvert q_n-q\rvert\mid\mathcal{F}_n]&=E_P[2(q_n-q)_++q-q_n\mid\mathcal{F}_n]\\ &=2E_P[(q_n-q)_+\mid\mathcal{F}_n]+E_P[q-q_n\mid\mathcal{F}_n]\\ &=2E_P[(q_n-q)_+\mid\mathcal{F}_n]\\ &\le2E_P[1_{\{q_n-q > \epsilon\}}(q_n-q)\mid\mathcal{F}_n]+2\epsilon\\ &\to2\epsilon \end{align} almost surely. As $\epsilon > 0$ can be taken arbitrarily small, the limit $$ E_P[\lvert q_n-q\rvert\mid\mathcal{F}_n]\to0 $$ (almost surely) follows.