Let $k$ be a $p$-adic field with integer ring $\mathcal{O}_k \subseteq k$, maximal ideal $m_k \subseteq \mathcal{O}_k$ and residue field $\mathbb{F}_q = \mathcal{O}_k/m_k$. Let $X$ be a smooth, geometrically integral scheme over $\mathcal{O}_k$ with good reduction mod $m_k$ and let $f: X \longrightarrow spec(\mathcal{O}_k[[t]])$ a surjective map whose generic fiber is smooth and geometrically integral (automatically with good reduction mod $m_k$). Assume that the fiber $X_0$ of $f$ over $t=0$ is a strict normal crossing divisor in $X$ whose geometric components $D_1,...,D_r$ are all reduced. Assume in addition that $X_0$ intersects the subscheme $X \otimes_{\mathcal{O}_k} \mathbb{F}_q$ transversely, so that $X_0 \otimes_{\mathcal{O}_k} \mathbb{F}_q$ is a strict normal crossing divisor in $X \otimes_{\mathcal{O}_k} \mathbb{F}_q$ whose geometric component are $D_1 \otimes_{\mathcal{O}_k} \mathbb{F}_q, ..., D_r \otimes_{\mathcal{O}_k} \mathbb{F}_q$. Let $a \in k$ be an element of positive valuation and let $P: spec(\mathcal{O}_k) \longrightarrow spec(\mathcal{O}_k[[t]])$ be the point corresponding to the map $t \mapsto a$. Let $x$ be an $\mathbb{F}_q$-point of $X_0 \otimes_{\mathcal{O}_k} \mathbb{F}_q$.
Claim: The following are equivalent:
1) There exists a $\mathcal{O}_k$-point $Q \in X(\mathcal{O}_k)$ which lies above $P$ and specializes to $x$.
2) The number of geometric components of $X_0 \otimes_{\mathcal{O}_v} \mathbb{F}_q$ which contain $x$ is at most the valuation of $a$.
Question: Is this claim true?
Let me describe a motivating example. Let $L$ be a finite dimensional unramified \'etale algebra over $k$ and let $\mathcal{O}_L \subseteq L$ be its ring of integers. Let $w_1,...,w_n$ be a basis for $\mathcal{O}_L$ as an $\mathcal{O}_k$-module and let $X \subseteq \mathbb{A}_{\mathcal{O}_k}^n \times spec(\mathcal{O}_k[[t]])$ be given by the equation $$ N_{L/k}(x_1w_1 + ... + x_nw_n) = t $$ with the obvious projection $f:X \longrightarrow spec(\mathcal{O}_k[[t]])$. The irreducible components of $X_0$ correspond to prime ideals of $\mathcal{O}_L$ which lie above $m_x$. An irreducible component corresponding to a prime ideal $I \subseteq \mathcal{O}_L$ splits into $[\mathcal{O}_L/I:\mathbb{F}_q]$ components over the maximal unramified extension of $k$. Hence the number of geometric components of $X_0$ is $\sum_{I | m_k} [\mathcal{O}_L/I:\mathbb{F}_q] = [L:k]$. In the notation above, we see that points $Q \in X(\mathcal{O}_k)$ which lies above $P$ correspond to elements $b \in \mathcal{O}_L$ whose norm is $a$. If $b \in \mathcal{O}_L$ is such an element (and $Q_b \in X(\mathcal{O}_k)$ is the associated point) then the number of geometric components of $X_0 \otimes_{\mathcal{O}_k}\mathbb{F}_q$ containing the reduction of $Q$ is exactly $n = \sum_{I | (b)}[\mathcal{O}_L/I:\mathbb{F}_q]$, and clearly $n \leq val(a)$. It is straightforward to verify that the other implication holds as well.
