New answer:
I'm going to explain in what sense the answer is yes, and why the "counterexample" I gave in my last answer was contrived.
First, recall that the cohomology $H^i(X, F) = R^i\Gamma(X, F)$ is only defined up to natural isomorphism, and depends on the choice of injective resolution of $F$. So, let's denote by $H^i(F,I)$ the cohomology with respect to an injective resolution $I$ of $F$.
If we pick a different resolution $J$ there is a natural map $H^i(F,I) \cong H^i(F,J)$ which doesn't depend on all the choices you make to construct it. So, if $I$ happens to be a resolution by $\mathcal O$-modules, this makes $H^i(F,I)$ an $\mathcal O$-module, and even if $J$ is not a resolution by $\mathcal O$-modules the natural isomorphism gives its cohomology an $\mathcal O$-module structure.
Since the map from the Cech sheaf to an injective resolution also gives natural maps $\check H^i(X,F) \to H^i(F,I)$ and $\check H^i(X,F) \to H^i(F,J)$, these will respect both the $\mathcal O$-action on $I$ and on $J$.
So in this sense the answer is yes, the map is always an isomorphism of $\mathcal O$-modules.
The example I gave originally is actually pretty contrived. It arose from this observation: If you take a complex of $\mathcal O$-modules $J$ which is an injective resolution of $F$ as abelian groups, but for which the augmentation $\varepsilon : F \to J$ is not $\mathcal O$-linear, then the map $\check H^i(X,F) \to H^i(F,J)$ will not be $\mathcal O$-linear, nor will the natural isomorphism $H^i(F,I) \cong H^i(F,J)$ with respect to be $\mathcal O$-actions.
Original answer:
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $\mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $\mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $\mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $\mathcal F = \mathcal O_X = \underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $\mathcal F$ with an injective resolution. One injective resolution is $\phi: \underline{k} \to \underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $\mathcal O$-linear by design because it coincides with $\phi$ itself.
So, if you want an $\mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $\Gamma(\mathcal O,X)$-linear.
Best Answer
Consider a good open cover, $(U_j)$ ($U_j$ is contractible, a finite intersection of $U_j's$ is contractible), and $g_{jk}:U_j\cap U_k\rightarrow U(1)$, there exist $h_{jk}:U_j\cap U_k\rightarrow \mathbb{R}=Lie(U(1))$ such that $g_{jk}=exp(ih_{jk})$, we have $g_{jk}g_{kl}=g_{jl}$ is equivalent to $exp(ih_{jk})exp(ih_{kl})=exp(i(h_{jk}+h_{kl}))=exp(ih_{jl})$, we deduce that $dh_{jk}$ is a cocycle since the sheaf of $1$-form is fine (there exist partition of unity). We deduce that there exists $1$-form $h'_{j}$ defined on $U_j$ such that $h'_k-h'_j=h_{jk}$, write $\omega$ the $2$-form whose restriction to $U_j$ is $dh'_j$.
Conversely, consider a closed $2$-form $\omega$. There exists a $1$-form $h'_j$ defined on $U_j$ such that $dh'_j=\omega_{\mid U_j}$, on $U_{jk}$, we have $dh'_k-dh'_j=0$, this implies there exists a function $h_{jk}$ such that $dh_{jk}=h'_k-h'_j$. Write $h_{jkl}=h_{kl}-h_{jl}+h_{jk}$ on $U_j\cap U_k\cap U_l$, $dh_{ijk}=0$, the fact that $\omega$ is a integral form implies that the constant form $h_{jkl}\in\mathbb{Z}$, we can write $g_{jk}=exp(ih_{jk})$ and $g_{jk}g_{kl}=g_{jl}$.