Working locally on $X$ and $Y$, we may assume they are affine and so the map $f : X \to Y$ corresponds to a ring map $S \to R$ (an inclusion) with $S$ smooth over the base field and $R$ normal. Then the generic fiber is simply $(S \setminus 0)^{-1} R$. That's certainly normal since a multiplicative set times a normal ring is still normal. It easily follows that an open set of the other fibers are also normal.

Working locally on $X$ and $Y$, we may assume they are affine and so the map $f : X \to Y$ corresponds to a ring map $S \to R$ (an inclusion) with $S$ smooth over the base field and $R$ normal. Then the generic fiber is simply $(S \setminus 0)^{-1} R$. That's certainly normal since a multiplicative set times a normal ring is still normal.

EDIT: If additionally the generic fiber is geometrically normal (this is not free, and may fail as Daniel points out above), then it easily follows that an open set of the other fibers are also normal.