Are typical sets larger, when information is messier?
Recall that
$$
A_\epsilon^{(n)}(P) = \left\{ (x_1,\ldots,x_n) \in \{0,1\}^n : 2^{-n(h(p)+\epsilon)} \leq p^{\sum_{i=1}^n x_i} (1-p)^{\sum_{i=1}^n (1-x_i)} \leq 2^{-n(h(p)-\epsilon)} \right\}.
$$
We can rephrase the condition as follows. First, it is equivalent to
$$
2^{-n(h(p)+\epsilon+\log(1-p))} \leq \left(\frac{p}{1-p}\right)^{\sum_{i=1}^n x_i} \leq 2^{-n(h(p)-\epsilon+\log(1-p))}.
$$
Taking the logarithm, we obtain
$$
-n(h(p)+\epsilon+\log(1-p)) \leq \log_2 \frac{p}{1-p} \sum_{i=1}^n x_i \leq -n(h(p)-\epsilon+\log(1-p)).
$$
This means that
$$
|A_\epsilon^{(n)}(P)| = 2^n \Pr\left[\frac{n(h(p)-\epsilon+\log(1-p))}{\log \frac{1-p}{p}} \leq \mathrm{Bin}(n,1/2) \leq \frac{n(h(p)+\epsilon+\log(1-p))}{\log \frac{1-p}{p}}\right]
$$
In order to estimate the probability on the right, we need to use binomial tail bounds. The Chernoff bounds is a well-known upper bound, and there are almost matching lower bounds, see for example (1) and (3) in Pelekis, A lower bound on binomial tails: an approach via tail
conditional expectations. Plugging these bounds, we would get a rather accurate estimate for $|A_\epsilon^{(n)}(P)|$, which we could then compare with $|A_\epsilon^{(n)}(Q)|$, estimated using the same method.

Unfortunately, this calculational method seems like a lot of work, better done with the aid of a computer algebra system, and I am not motivated enough to actually carry it through. However, it seems likely (if not certain) that this method will be able to settle the conjecture.