Let $O$ be a valuation ring with fraction field $K$ of characteristic zero and residue field $O/m=k$ of characteristic $p>0$, and $X$ be a proper smooth scheme over $O$. Then can we control the mod $p$ etale cohomology of the special fiber by that of the generic fiber? Namely, do we have $\dim_{\Bbb F_p}H^i_{et}(X_k,\Bbb F_p) \leq\dim_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p)$ in general?

I am interested in the case $O=O_{\Bbb C_p}$, and it turns out mod p etale cohomology vanish for $i>\text{dim}(X_k)$ by Artin-schreier sequence while the $2\text{dim}(X_K)$'s etale cohomology can be nonzero for the generic fiber, see this answer. So they are not of the same dimension in general and it's natural to ask for a bound.

Also, smooth base change theorem is good for $\Bbb Z/\ell$ ($\ell \not=p$) but don't hold if the coefficient sheaf has $p$-torion. From BMS we know $\text{dim}_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p) \leq \text{dim}_{k}H^i_{dR}(X_k)$, so maybe a lower bound is also possible in our setting.

Edit: The case $i=0,1$ is true. What about the case $i=2$? Moreover, $\text{dim} _{\Bbb F_p}H^i_{et}(X,\Bbb F_p)$ is finite and less than $\text{dim}_k H^i(X_k, O_k)$, see Lemma 0A3L for a related semi-linear algebra result (so maybe we need to understand the Frobenius action).

Edit: Note that $LHS \leq \text{dim}_{k}H^i_{k}(X_k,O_{X_k})\geq \text{dim}_{C}H^i(X_C,O_{X_C}) \leq \text{dim}_{C}H^i_{sing}(X_K^{an},K) \leq RHS. $ The only $\geq$ is due to upper semi-continuities, and we use an abstract isomorphism between $\Bbb C$ and $K$ to apply singular cohomologies, Hodge-de Rham decomposition and the universal coefficient theorem. So if $\text{dim}_{k}H^i_{k}(X_k,O_{X_k})=\text{dim}_{C}H^i(X_C,O_{X_C})$ then what we want holds, however this is not true in general. For example, one considers lifting of a singular Enriques surface in char $2$, but even in this case the inequality still holds (because the second Betti number is $10$, which is much larger than $h^{2,0}=1$).