Smooth projective surface with geometrically integral reduction

That is not true. There are probably shorter answers than the following. Let $K$ be a field, and denote a separable closure by $K^{\text{sep}}$. Let $n>1$ be an integer.

Definition. A Severi-Brauer variety over $K$ of relative dimension $n-1$ is a proper, smooth $K$-scheme whose base change to $K^{\text{sep}}$ is isomorphic to projective space of dimension $n-1$ over $K^{\text{sep}}$.

There is a natural bijection between the set of $K$-isomorphism classes of Severi-Brauer varieties over $K$ of relative dimension $n-1$ and the subset of $\text{Br}(K)[n]$ of those ed. $n$-torsion elements in the Brauer group of $K$ whose order index divides $n$. In particular, the identity element in this subset corresponds to the isomorphism class of projective space, i.e., the isomorphism class of any Severi-Brauer variety over $K$ of relative dimension $n-1$ that has a $K$-rational point.

For a finite, separable field extension $L/K$ of degree $d$, there are restriction and corestriction group homomorphisms, $$ \text{Br}(K)\to \text{Br}(L), \ \ \text{Br}(L)\to \text{Br}(K), $$ whose composition is the "multiplication by $d$" map.

Finally, for every finite extension of $\mathbb{Q}_p$, local class field theory gives a natural isomorphism, $$ \text{inv}_K:\text{Br}(K) \to \mathbb{Q}/\mathbb{Z}, $$ ed. and the index of every element equals the order of that element.

Now let $K$ be a finite extension of $\mathbb{Q}_p$, and let $X_K$ be a Severi-Brauer variety over $K$ of relative dimension $n-1$ whose image in $\text{Br}(K)[n]$ is a generator for this cyclic group of order $n$. By the restriction-corestriction homomorphisms, for a finite field extension $L/K$, the base change of $X_K$ over $L$ has an $L$-rational point only if the degree $d$ of the field extension is divisible by $n$.

On the other hand, if $X_K$ has a proper, flat model over the ring of integers of $K$ whose special fiber has nonempty smooth locus an irreducible component that is geometrically integral, then by the Lang-Weil estimates together with Hensel's Lemma, for every sufficiently large integer $d$, there is an unramifield field extension $L/K$ of degree $d$ such that the base change has an $L$-rational point. Therefore, there is no proper, flat model of $X_K$ over the ring of integers of $K$.

In particular, for the integer $n=3$, there exists a Severi-Brauer scheme over $\mathbb{Q}_p$ of relative dimension $n-1=2$ that has no proper, flat model over $\mathbb{Z}_p$ (ed. . . . with an irreducible component that is geometrically integral!).

Let $X$ be an integral regular scheme which is proper over $\mathbb{Z}_p$. Assume that the special fibre $X_{\mathbb{F}_p}$ is irreducible and let $k$ be the algebraic closure of $\mathbb{F}_p$ in the function field of $X_{\mathbb{F}_p}$. Then $k$ is a birational invariant of the generic fibre $X_{\mathbb{Q}}$. (The field $k$ is something like the field of definition of the geometric irreducible components of $X_{\mathbb{F}_p}$, with $k = \mathbb{F}_p$ if and only if $X_{\mathbb{F}_p}$ is geometrically irreducible).

This is a special case of a more general result given in [1, Cor. 2.3] (which also makes precise what I mean by "birational invariant of the generic fibre", also allows the special fibre to be reducible, and applies to general dvrs).

So consider the $\mathbb{Z}_p$-scheme

$$ x^2 - ay^2 = p z^2$$

where $a \in \mathbb{Z}_p^\times$ is a non-square (model of a plane conic). This is easily check to be integral regular and proper. But the special fibre is

$$x^2 - ay^2 \equiv 0 \bmod p$$ which has $k = \mathbb{F}_p(\sqrt{a})$ in the above notation. Therefore, by the above result, there is no regular integral proper model of the generic fibre with geometrically integral special fibre.

For a counter-example involving surfaces, just take the fibre product of the above scheme with $\mathbb{P}^1_{\mathbb{Z}_p}$.

Your question allowed arbitrary (not necessarily regular) models. If you are given a non-regular model $Y$ for the above conic, then you can perform resolution of singularities to obtain a regular model $\widetilde{Y}$ (since $\dim Y \leq 3$). If $Y$ had a geometrically integral special fibre, then the special fibre of $\widetilde{Y}$ would have an irreducible component which is geometrically integral, which is not allowed by the more general [1, Cor. 2.3].

[1] Skorobogatov - Descent on toric fibrations