The skier has the following forces acting on him:
- a normal force $F_N$
- a gravitational force $F_W=mg$
- a force from air friction that is probably well approximated as $F_a=cAv^2$, where $A$ is the skier's cross-sectional area
- a frictional force from the snow $F_f$
$F_a$ increases with speed, and $F_f$ may also have some velocity dependence as well, and this is why a terminal velocity exists. The terminal velocity is the one at which the vector sum of these forces is zero.
If $F_f$ is negligible, then clearly a heavier skier will have a greater $v$, because we need a bigger $F_a$ to cancel out the component of $F_W$ parallel to the slope.
If $F_f$ obeys the usual freshman-physics model of kinetic friction, then it's velocity-independent, and a simple calculation shows that the same conclusion holds for fixed $A$: a heavier skier goes faster. In reality, a bigger skier has a bigger $A$ as well. However, I don't think the qualitative result is altered, because under geometrical scaling, $m$ grows faster than $A$.
However, I don't think it's necessarily valid to assume that $F_f$ obeys the usual model of kinetic friction, which is usually found to be valid only when a solid, dry surface slides over another solid, dry surface. Here the skier is leaving a track by pushing aside the snow (unless the conditions are very icy). I would guess that something like $F_f\propto F_N v$ holds. If this form of $F_f$ applies and $F_a$ is absent, then all forces are proportional to mass, and the terminal velocity is independent of mass. If this form of $F_f$ exists along with an $F_a$, then I think you'd come back to the same result, which is that heavier skiers would go faster.
But the whole thing really depends on the model of friction, and I don't think any one model of friction will be accurate for a variety of conditions (ice, deep powder, ...).