$\newcommand{\F}{\mathcal{F}}$The answer is yes.
Indeed, by time rescaling, without loss of generality (wlog) $r=1$.
Take any random variable (r.v.) $Z$ with $EZ=0$, and let $Z'$ be an independent copy of $Z$. Then, by Jensen's inequality, for any real $z$ we have $E|Z|\le E|Z-Z'|=E|(Z-z)-(Z'-z)|\le2E|Z-z|$, so that
\begin{equation*}
E|Z|\le2E|Z-z|. \tag{1}
\end{equation*}
Letting $E_t:=E(\cdot|\F_t)$ and using (1) with $Z=X_t-X_{t-1}$ and real $t\ge1$, we get
\begin{equation*}
\begin{aligned}
E_{t-1}|X_t-X_{t-1}|&\le2E_{t-1}|(X_t-X_{t-1})-(Y_t-X_{t-1})| \\
&=2E_{t-1}|X_t-Y_t|,
\end{aligned}
\end{equation*}
since $Y_t-X_{t-1}$ is $\F_{t-1}$-measurable.
So, $E|X_t-X_{t-1}|\le2|X_t-Y_t|$. So, for any real $u\ge0$ and any natural $T$,
\begin{equation*}
S_u:=\sum_{n=T}^\infty E|X_{n+u}-X_{n-1+u}|\le2\sum_{n=T}^\infty E|X_{n+u}-Y_{n+u}|
\end{equation*}
and hence
\begin{equation*}
\begin{aligned}
\int_0^1 du\, S_u &\le2\sum_{n=T}^\infty \int_0^1 du\, E|X_{n+u}-Y_{n+u}| \\
& =2\int_T^\infty dt\, E|X_t-Y_t|<\infty
\end{aligned}
\end{equation*}
if $T$ is large enough, by your displayed condition. So, $S_u<\infty$ for some $u\in[0,1]$. By time shift, wlog $u=1$. So,
\begin{equation*}
\sum_{n=T}^\infty E|X_{n+1}-X_n|=S_1<\infty. \tag{2}
\end{equation*}
So, $(X_n)$ converges in $L^1$ and hence $(E|X_n|)$ is bounded. So, by Doob's martingale convergence theorem,
\begin{equation*}
X_n\to Y \tag{3}
\end{equation*}
as $n\to\infty$ almost surely (a.s.) for some (real-valued) r.v. $Y$.
Next, by Doob's martingale inequality,
\begin{equation*}
P(\max_{t\in[n,n+1]}|X_t-X_n|>h)\le\frac{E|X_{n+1}-X_n|}h
\end{equation*}
for any real $h>0$.
So, by the Borel–Cantelli lemma and (2), for each real $h>0$ a.s. there will be only finitely many natural $n$ such that $\max_{t\in[n,n+1]}|X_t-X_n|>h$. That is, $\max_{t\in[n,n+1]}|X_t-X_n|\to0$ a.s. as $n\to\infty$.
Thus, by (3), $X_t\to Y$ a.s. as $t\to\infty$, as claimed.
Best Answer
I think that I have a counterexample. Let $(X,Y)$ be a Brownian motion in $\mathbb{R}^2$. Then $M = \int_0^\cdot X_s \mathrm{d}Y_s$ is a martingale, in the natural filtration of $(X,Y)$, in its own filtration $(\mathcal{F}_t)_{t \ge 0}$ and also in $(\mathcal{F}_t \vee \sigma(X))_{t \ge 0}$. Yet, $X$ is not independent of $M$ since $\langle M \rangle = \int_0^\cdot X_s^2\mathrm{d}s$ is not deterministic.
Remark: in this example, $\mathcal{F}$ is not immersed in $\mathcal{H}$ since $X$ is no more a martingale in $\mathcal{H}$.