The answer to my question is **yes!!** [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): [On the mean ergodic theorem for subsequences](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-66/issue-4/On-the-mean-ergodic-theorem-for-subsequences/bams/1183523588.full)." I now give the proof of the answer to my question, in my own words.
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First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

**Definition.** We say that $f_n$ **converges uniformly modulo re-ordering to** $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$,
$$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

**Proposition.** *Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence*:
$$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$
*Then $f_n$ converges uniformly modulo re-ordering to $f$.*

*Proof.* Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$
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Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by
\begin{align*}
f_n(B) &= \mu(A \cap T^{-n}(B)) \\
f(B) &= \mu(A)\mu(B).
\end{align*}

**Theorem.** *$f_n$ converges uniformly modulo re-ordering to $f$.*

We now prove this, using the above Proposition. We will equip the space of $[0,1]$-valued functions on $X$ identified up to $\mu$-almost sure equality with its natural topology, namely the topology of $L^p$ convergence for any $p \in [1,\infty)$ or equivalently the topology of convergence in probability.

Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$,
\begin{align*} f_n(B) &= \int_{T^{-N}(B)} \mathbf{1}_{T^{n-N}(A)} \, d\mu \\
f(B) &= \int_{T^{-N}(B)} \mu(A) \, d\mu.
\end{align*}
So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives
$$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{T^{-\max J}(B)} \mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \, d\mu. $$
Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

**Lemma.** *Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence*:
$$ \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \to \mu(A) \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

*Proof.* We have
\begin{align*}
\| \mathrm{RHS}-\mathrm{LHS} \|_{L^2(\mu)}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_{T^{n-\max J}(A)}) \right\|_{L^2(\mu)}^2 \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X (\mu(A) - \mathbf{1}_{T^{n_1-\max J}(A)})(\mu(A) - \mathbf{1}_{T^{n_2-\max J}(A)}) \, d\mu \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X \mu(A)^2 - \mu(A)\mathbf{1}_{T^{n_2-\max J}(A)} - \mu(A)\mathbf{1}_{T^{n_1-\max J}(A)} + \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_X \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu - \mu(A)^2 \right) \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\
&\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square
\end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]