Let $K$ be a finite extension of $\mathbb{Q}_p$ with the ring of integers $\mathcal{O}_K$ and the residue field $k$.
By a theorem of Berthelot and Ogus(https://link.springer.com/article/10.1007%2FBF01389319 -- unfortunately, I can't find a freely acessible version), for any smooth proper scheme $X$ over $\mathcal{O}_K$ there is an isomorphism $$H^i_{cris}(X_k/W(k))\otimes_{W(k)}K\simeq H^i_{dR}(X/\mathcal{O}_K)\otimes_{\mathcal{O}_K}K \qquad \qquad(*)$$
Of course, if the ramification index $e$ of $\mathcal{O}_K$ is less or equal to $p-1$ then the integral version of this isomorphism holds $$H^i_{cris}(X_k/W(k))\otimes_{W(k)}\mathcal{O}_K\simeq H^i_{dR}(X/{\mathcal{O}_K})\qquad\qquad\qquad (**)$$ as $\mathcal{O}_K\to k$ is a (pro) divided power thickening. But there seems to be no reason to expect such isomorphism in general. It is even shown in Remark 2.10 in the paper above that $(*)$ does not carry the $\mathcal{O}_K$-lattice on the left to the one on the right. I wonder if there is an explicit way to see that $(**)$ cannot possibly hold in general.
Is there an example of $X/\mathcal{O}_K$ such that torsions in $H^i_{cris}(X_k/W(k))\otimes_{W(k)}\mathcal{O}_K$ and $ H^i_{dR}(X/{\mathcal{O}_K})$ are different?
For instance, any $X$ which has in $H^i_{dR}(X/\mathcal{O}_K)$ a direct summand of length not divisible by $e$ would be an example as a module of the form $M\otimes_{W(k)}\mathcal{O}_K$ can't have such.
A refined version of $(*)$ seems to be provided by the Theorem 1.2 in https://arxiv.org/pdf/1802.03261.pdf and, as far as I understand, it does not prohibit the torsions from being different. E. g., suppose that $K=\mathbb{Q}_p(p^{1/p^2})$ and $H^i_{\mathfrak{S}}(X)$ is the Breuil-Kisin module given by $\mathbb{F}_p$ ($u\in \mathfrak{S}=\mathbb{Z}_p[[u]]$ acts by zero and Frobenius is zero or identity) while $H^{i+1}_{\mathfrak{S}}(X)$ is torsion-free. Then $H^i_{dR}(X/\mathcal{O}_K)=\mathbb{F}_p\otimes_{\mathbb{Z}_p[[u]],u\mapsto p^{p/p^2}}\mathbb{Z}_p[p^{1/p^2}]=\mathbb{F}_p[t]/t^p$ and $H^i_{cris}(X_k/W(k))=\mathbb{F}_p\otimes_{\mathbb{Z}_p[[u]]}\mathbb{Z}_p=\mathbb{F}_p$ but $\mathbb{F}_p\otimes_{\mathbb{Z}_p}\mathcal{O}_K=\mathbb{F}_p[t]/t^{p^2}$
