Yes I think so. We may assume that $\mathcal F$ is the $\sigma$-algebra generated by sets of the form $X_n^{-1}(a,\infty)$ as $n$ runs over the positive integers and $a$ runs over the reals as $\mathcal F$ already contains these sets. In so doing, we might be making $\mathcal F$ smaller, but that only makes the problem harder.
Let $\mathcal P_1,\mathcal P_2,\mathcal P_3,\ldots$ be a sequence of refining finite partitions of $\mathbb R$ such that the intersection of an element of $\mathcal P_n$ for each $n$ consists of at most one point. For example, $\mathcal P_n$ could consist of intervals $[n,\infty)$, $(-\infty,-n)$ and half-open dyadic intervals of length $1/2^n$ between $-n$ and $n$.
Then let $\mathcal F_n$ be the finite sub-algebra of $\mathcal F$ generated by
the sets $X_i^{-1}A$ for $1\le i\le n$ and $A$ running over $\mathcal P_n$.
Write $\mathcal F_n=\{B^n_1,\ldots B^n_{k_n}\}$. Now there is a measurable map $\Phi$ from $\Omega$ to $\Xi=\prod_{n=1}^\infty\{1,\ldots,k_n\}$ sending $\omega$ to sequence of partition elements that it lies in.
Equip $\Xi$ with the lexicographic ordering. We can then check that $\mathbb P$ induces a measure $\mu$ on $\Xi$. There are at most countably many atoms in $\Omega$ with respect to $\mathcal F$. Let the set of atoms be $A$. We can map each of these atoms disjointly to atoms of the same mass (at least) in the limit measure $\tilde {\mathbb P}$ (some justification needed here, but I'm pretty confident).
First we couple the atoms in both $\tilde{\mathbb P}$ and $\Omega$. That is we define $X$ restricted to the atoms.
To finish, use quantile coupling to couple what's left. Given an $\omega$, set $t(\omega)=\mu(\{\omega'\in \Omega\setminus A\colon \Phi(\omega')\le \Phi(\omega)\})$ and finally, define for $\omega$ not in an atom, $X(\omega)=\inf\{s\colon \big(\tilde{\mathbb P}(-\infty,s])-\sum\text{(atoms of $\tilde{\mathbb P}\le s$)} \big)\ge t\}$.
