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." I now give the proof of the answer to my question, in my own words.
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 $k$. $\ \square$
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.]