Can Hodge symmetry fail if there is a lift to $W_2$ and the crystalline cohomology is torsion-free? Let $f:X\to \mathrm{Spec}\:\mathbb{F}_p$ be a smooth proper morphism with $p>\mathrm{dim}\:X$. Assume that $H^i_{\mathrm{crys}}(X/\mathbb{Z}_p)$ is torsion-free for all $i\geq 0$ and that there is a proper flat morphism $X_2\to \mathrm{Spec}\:\mathbb{Z}/p^2\mathbb{Z}$ that reduces to $f$. Does it follow that $\mathrm{dim}_{\mathbb{F}_p} H^i(X, \Omega^j_{X/\mathbb{F}_p})=\mathrm{dim}_{\mathbb{F}_p} H^j(X, \Omega^i_{X/\mathbb{F}_p})$ for all $i, j\geq 0$?
If we assume that $H^i(X, W\Omega^j_X)$ are finite $\mathbb{Z}_p$-modules for all $i, j\geq 0$ then it follows from theorems of Joshi and Deligne-Illusie. If we assume that there is a proper flat morphism $X_{\infty}\to \mathrm{Spec}\:\mathbb{Z}_p$ that reduces to $f$ then it follows from the universal coefficient formula and Hodge symmetry in characteristic 0 (first established by Deligne).
 A: There are counterexamples (at least for some $p$) even if we assume that $X$ lifts all the way to a (non-algebraizable) formal scheme over $\mathbb{Z}_p$. See e.g. Theorem 4.1 in https://arxiv.org/pdf/2005.02226.pdf
This example is obtained by taking a quotient of a formal abelian scheme $A$ by a free action of a finite group of order prime to $p$ to get a formal scheme $\mathfrak{X}$ over $\mathbb{Z}_p$ so that $H^i(\mathfrak{X},\Omega^j_{\mathfrak{X}})$ is the submodule of invariants in $H^i(A,\Omega^j_A)$. For any pair of degrees $i+j\geq 3, i\neq j$ it can be arranged (if we also further take the product with an appropriate blow-up of a complete intersection) that $$rk \,H^i(\mathfrak{X},\Omega^j_{\mathfrak{X}})\neq rk\, H^j(\mathfrak{X},\Omega^i_{\mathfrak{X}})$$ Since all Hodge cohomology groups are free, the universal coefficients formula gives $H^i(X,\Omega^j_{X/\mathbb{F}_p})=H^i(\mathfrak{X},\Omega^j_{\mathfrak{X}})\otimes_{\mathbb{Z}_p}\mathbb{F}_p$ for $X:=\mathfrak{X}\times_{\mathbb{Z}_p}\mathbb{F}_p$, so the Hodge symmetry for $X$ fails as well. The crystalline cohomology of $X$ are free modules because they are similarly equal to the invariants of the group action on the crystalline cohomology of an abelian variety.
The requirement $p>\dim X$ is slightly subtle to fulfill as the dimension of $X $ in the construction depends on an auxiliary prime $l$, but at least for $p$ congruent to $2$ or $3$ mod $5$ we can arrange $X$ to be $5$-dimensional (see Remark 3.11(ii) in the linked paper) which gives examples for all primes $p>5$ with such residues.
