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Here is an example showing the answer is no:

Start with $Z =\mathbb{P}^2_R$, $R$ an arbitrary dvr. Let $P$ be a section of $Z \to Spec(R)$ and let $W$ be the blowup of $Z$ along the image of the section (so both fibres are $\mathbb{P}^2$ with a point blown up). Let $Q$ be a section of $W \to Spec(R)$ whose image does not intersecting intersect the exceptional divisor of the first blow up and let $R$ Q'$be a section which whose image intersects the exceptional divisor only in the special fibre. Let$X$be the blow up of$W$along$Q$and$Y$the blow up of$W$along$R$.Q'$.

The generic fibres of $X$ and $Y$ are isomorphic since they are both just $\mathbb{P}^2$ over some field blown up in two distinct (rational) points and any two points on $\mathbb{P}^2$ over a field are "the same". That the special fibres are not isomorphic can be seen by considering curves of self intersection $-2$:

The special fibre of $X$ has none since we have blown up two distinct points, so the only curves with negative self intersection are three $(-1)$-curves. (-1)$-curves i.e. the two exceptional divisors and the strict transform of the line joining the two points. The special fibre of$Y$does have one; this is the strict transform in the second blowup of the exceptional divisor of the first blowup. (The second blow up changes the$(-1)$-curve into a$(-2)$-curve since we blow up a point on it.) 1 Here is an example showing the answer is no: Start with$Z =\mathbb{P}^2_R$,$R$an arbitrary dvr. Let$P$be a section of$Z \to Spec(R)$and let$W$be the blowup of$Z$along the image of the section (so both fibres are$\mathbb{P}^2$with a point blown up). Let$Q$be a section of$W \to Spec(R)$not intersecting the exceptional divisor of the first blow up and let$R$be a section which intersects the exceptional divisor only in the special fibre. Let$X$be the blow up of$W$along$Q$and$Y$the blow up of$W$along$R$. The generic fibres of$X$and$Y$are isomorphic since they are both just$\mathbb{P}^2$over some field blown up in two distinct (rational) points and any two points on$\mathbb{P}^2$over a field are "the same". That the special fibres are not isomorphic can be seen by considering curves of self intersection$-2$: The special fibre of$X$has none since we have blown up two distinct points so the only curves with negative self intersection are three$(-1)$-curves. The special fibre of$Y\$ does have one; this is the strict transform in the second blowup of the exceptional divisor of the first blowup.