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.
@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."
Best Answer
This is just a comment on your definition of "uniformly recurrent". Can you give an example of a system you'd consider uniformly recurrent?
If this notion is standard, the following must be nonsense, but it seems to me that your notion does not make much sense.
Proof. Suppose first that the measure of points with an eventual period is $0$. Then by the Rokhlin lemma for noninvertible transformations [1], there exists an $n$-tower for $f$ of measure at least $1 - \epsilon$ for each $\epsilon > 0$, meaning a set $B_{(n)}$ such that for some base $B$, $B_{(n)} = \bigcup_{0 \leq k < n} T^{-k}(B)$ is disjoint and has measure at least $1 - \epsilon$ and the sets $f^{-1}_k(B)$ are disjoint.
From this it is easy to see that there also exist arbitrarily small such sets, just take small subsets of $B$. From this we obtain that we can find for each $n \geq 1$ pick an $n$-tower $B_{(n)}$ such that these towers are disjoint for distinct $n$. Now denoting by $B^n$ is the base of $B_{(n)}$, define $A = \bigcup_{n > 1} B^n \cup T^{-n+1}(B^n)$.
Note that $B^n$ and $T^{-n+1}(B^n)$ have equal measure, so if you pick a random point $x$ in $A$, it will be in in a set of the form $T^{-n+1}(B^n)$ (for some $n$) with probability $0.5$. But then the minimal $n$ of $T$ such that $T^n(x) \in A$ is $n-1$. It follows that $T$ is not uniformly recurrent.
We conclude that there must exist $n, p$ such that the probability that $T^n(x) = T^{n+p}(x)$ is positive. In particular, the shift-invariant set of points satisfying $T^p(x) = x$ has positive measure, so this measure must be $1$. It is well-known that in this case, we can find a cross-section $S$ such that $X = S \cup T(s) \cup \cdots T^{n-1}(S)$ and the union is disjoint (assuming this is on a standard Lebesgue space $[0, 1]$, I think the usual construction is to pick the minimal point on each orbit).
Ergodicity implies that $S$ must be a singleton, as otherwise we can split it in two. Square.
Of course in this case, the answer to your question is "yes".
[1] Avila, Artur; Candela, Pablo, Towers for commuting endomorphisms, and combinatorial applications, Ann. Inst. Fourier 66, No. 4, 1529-1544 (2016). ZBL1360.28012.