I would like to follow up on my previous comment. If indeed $\mathbb{D}(A)_R=\mathbb{D}(A) \otimes_{W(k)}R$, then there is a big problem. Let $f:A\rightarrow A$ be an isogeny which factors by multiplication by $p$: $f=pg$, where $g$ is another isogeny. Lifting $f$ to $R$ is the same as lifting in a compatible way with $f$ two Hodge filtrations. But the map induced by $f$ on $\mathbb{D}(A)_R$ is just the zero map because $f$ factors by $p$. So the compatibility condition disappears completely, and lifting $f$ reduces to lifting two copies of $A$. That looks unreasonable to me.

There's something I don't quite understand: To an abelian variety over $k=\overline{\mathbb{F}_p}$, we can associate a Dieudonné module $\mathbb{D}(A)$ which is a free $W(k)$-module. Now, according to Grothendick-Messing theory, for every locally nilpotent divided power extension $R/k$, there is also a free $R$-module $\mathbb{D}(A)_R$. What is the connection between $\mathbb{D}(A)$ and $\mathbb{D}(A)_R$ ? My first thought was that $\mathbb{D}(A)_R=\mathbb{D}(A)\otimes_{W(k)} R$, where $R$ is considered a $W(k)$-module via the map $W(k)\rightarrow k \rightarrow R$. Is this true ?

Yes, the map induced on $\mathbb{D}(A)$ is multiplication by $p$, but what about the map induced on $\mathbb{D}(A)_R$ ?