I follow the book of Effros+Ruan (which is a book, so not viewable online, but really is the nicest source I think). For any operator spaces $X,Y$ we can consider the operator space projective tensor product $\newcommand{\proten}{\widehat\otimes}X\proten Y$ whose dual satisfies
$$ (X\proten Y)^* = CB(X,Y^*). $$
(To be precise, there is some completely isometric isomorphism here.) This is in Chapter 7, Corollary 7.1.5 to be precise.

The Fubini tensor product is also considered in Chapter 7, Theorem 7.2.3, which shows that $(X\proten Y)^* \cong X^* \bar\otimes_{\mathcal F} Y^*$ in general, so that $X^* \bar\otimes_{\mathcal F} Y^* \cong CB(X,Y^*)$.

So now apply this to a von Neumann algebra $M$ with predual $M_*$, and a dual operator space $X$ with predual $X_*$. Then
$$ CB(M_*, X) = (M_* \proten X_*)^* = M \bar\otimes_{\mathcal F} X. $$
Again there are (canonical) isomorphisms involved here, but chasing them down will show that they match the isomorphism given in the original question. In particular, this includes the isomorphism $M \bar\otimes_{\mathcal F} X \cong X \bar\otimes_{\mathcal F} M$.

Here I used that $X$ is a dual space, and in Chapter 7 of Effros and Ruan, we need this, and a "dual realisation" of $X$ as a weak$^*$-closed subspace of $B(H)$. That is, the Original Question has a positive answer when $X \subseteq B(H)$ is weak$^*$-closed.

When $X$ is only assumed norm closed, we can use the definition of the Fubini tensor product given in the original question, though I am not aware of much study of this. However, let $\overline{X}^{w^*}$ be the weak$^*$-closure of $X$ in $B(H)$. Then by definition(s),
$$ X \bar\otimes_{\mathcal F} M \subseteq \overline{X}^{w^*} \bar\otimes_{\mathcal F} M \cong CB(M_*, \overline{X}^{w^*}). $$
The isomorphism clearly identifies $X \bar\otimes_{\mathcal F} M$ with those CB maps $T:M_*\rightarrow \overline{X}^{w^*}$ which map into $X$, and by definition of what a CB map is, this is just the space $CB(M_*,X)$. Thus the original question is answered in the affirmative.