The Chacon transformation has a nice and fairly explicit description as a uniquely ergodic subshift: set $B_0=0$ and set $B_{k+1}=B_kB_k1B_k$.
The subshift is the set of all infinite words, all of whose finite subwords occur as a someword of some $B_k$.
From this it is easy to see why the lengths $n=|B_k|$ fail to be good times for strong mixing. As mentioned previously Parry shows the weak mixingness.
Reposting some proofs I gave in the comments as a (partial) answer, demonstrating that the answer to Julian's question is "yes" for Bernoulli/i.i.d. systems, and so that it is not resolved by some usual "there's no way to guarantee a rate of mixing" examples. The question actually seems pretty subtle, although my honest guess is that the answer is "no" in general.
Say $\mu$ is i.i.d. on $S^{\mathbb{Z}}$ for some finite set $S$ (and that $T = \sigma$ is the left shift on $S^{\mathbb{Z}}$), that $A$ depends only on coordinates $-k$ through $k-1$, and $\epsilon > 0$. Assume for a contradiction that Julian's property fails. Then, for all $n$, there exist $B_n$ and $i_0,...,i_{n-1}$ s.t. $|\mu(A\cap \sigma^{i_j}B_n)-\mu(A)\mu(B_n)| \geq \epsilon$ for $0 \leq j < n$. (I've removed the negative power for simplicity, since $\sigma$ is invertible here.)
We can assume without loss of generality that $i_0 > k$ and $i_j - i_{j-1} > 2k$ for all $0 < j < n$ (since this decreases the size of $\{i_0, \ldots, i_{n-1}\}$ by at most a factor of $2k$).
Note that $\mu$ is i.i.d., so exchangeable. Take a self-map $\pi_n$ of
$S^{\mathbb{Z}}$ sending coordinate $i_j + m$ to $2kj + m$ for $0 \leq j < n$ and $-k \leq m < k$, and consider $C_n=\pi_n(B_n)$; clearly $\mu(C_n) = \mu(B_n)$. Then, since $A$ depends only on coordinates in $[-k, k)$, $\mu(A \cap \sigma^{i_j} B_n) = \mu(A \cap \sigma^{2kj} C_n)$ for $0 \leq j < n$. So, $\mu(A\cap \sigma^{2kj} C_n)-\mu(A)\mu(C_n))| \geq \epsilon$ for $0 \leq j < n$.
Take a weakly convergent sequence $C_{m_n} \rightarrow C$. (Here, I mean that we identify a set $D$ with the measure $\mu_D$ defined by $\mu_D(E) = \mu(D \cap E)$, and take a weak(star) convergent subsequence.) Then, since $A$ is clopen (i.e. $\chi_A$ is continuous), $\mu(A\cap \sigma^{2kj}C)-\mu(A)\mu(C)| =
\lim_{n \rightarrow \infty} \mu(A\cap \sigma^{2kj}C_{m_n})-\mu(A)\mu(C_{m_n})| \geq \epsilon$ for all $j$. And this contradicts mixing of $\mu$.
Finally, we note that for arbitrary $A$ and $\epsilon$ (i.e. $A$ doesn't depend on only finitely many coordinates), there exists $A'$ where $\mu(A \triangle A') < \epsilon/3$. Then, for any set $B$, $|\mu(A)\mu(B) - \mu(A')\mu(B)| < \epsilon/3$, and for any $j$, $\mu(A\cap \sigma^j B) - \mu(A' \cap \sigma^j B)| < \epsilon/3$. By the triangle inequality, $|\mu(A' \cap \sigma^j B) - \mu(A')\mu(B)| < \epsilon/3 \Longrightarrow |\mu(A \cap \sigma^j B) - \mu(A)\mu(B)| < \epsilon$. This means that, using terminology from the original question, $N_{A, \epsilon} \leq N_{A', \epsilon/3}$. The second quantity has been proved finite, so the first is as well, completing the proof.
Best Answer
@John Griesmer answered this question: "I don't think there is a system that satisfies your definition of uniformly weak mixing. Such a system must be ergodic, and therefore admit Rohlin towers. Given $𝑁>0$ and a Rohlin tower $\{𝐸,𝑇𝐸,…,𝑇^{𝑚−1}𝐸\}$ with $𝑁=𝑜(𝑚)$, you can set $𝐴=𝐵=𝐸\cup 𝑇𝐸 \cup \ldots \cup 𝑇^{𝑚/2}𝐸$ and find that $\frac{1}{𝑁+1}\sum^{𝑁+1}_{𝑘=1}\mu(𝑇^{−𝑘}𝐴\cap 𝐴)\approx \mu(𝐴)\approx 1/2$. In other words, given an initial segment of integers, every ergodic MPS on a nonatomic probability space admits a subset which is approximately invariant under that segment."