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.)
