Do heavier skates mean less air?

This depends on if you are getting an air from a ramp, or by jumping. If you simply use a ramp, then the weight of your skates will not limit the height you get. This is determined by the speed and angle at which you leave the ramp.

In order for the skater to get height, their kinetic energy is converted into potential energy due to gravity (assuming conservation of energy)

1/2mv^2 = mgh

m = total mass of skater
v = component of velocity in the vertical direction
g = acceleration due to gravity
h = maximum height gained

        v^2
so, h = ---
        2g

i.e. mass is of no consequence. the only thing you can change to increase the maximum height attained is the vertical speed. This can be by skating faster, or by making the launch steeper (up to vertical) so that none of the speed is 'wasted' in horizontal movement.

If we take into account air resistance, then this is an extra force which is dependent on the shape of the skater and the velocity. It is always opposing the motion, so this force is independent of mass also. The deceleration due to this force is dependent of mass as:

a = F/m

F = force caused by air resistance
m = total mass of skater
a = acceleration opposing velocity (slowing down)

So as the skater's mass increases, they are slowed less by air resistance at the same speed (assuming the shape is the same)

A balloon skater and a lead skater leave the floor at the same speed and angle. Which will travel further? Without air present, they would travel the same distance in the same trajectory. With air present, the balloon skater would not go as far or as high. As the force due to drag is -Knv, the acceleration associated with this is definced as:

    -Knv
a = ----
     m

Where (SI units used):
K = Drag coefficient of the shape (about 18 Ns/m^2 for a 1m sphere)
n = coeffiecient of viscosity (about 0.000018 kg/ms for air)
v = velocity (maximum of about 6 m/s for a 2 metre high jump)
m = mass (about 60 kg for average aggro skater)

a = 0.14 m/s^2 maximum = 1.5% of accelleration due to gravity. Remember, this is the maximum, and decreases rapidly as the height increases and speed decreases. I'd call this negligable.

However, can't this formula be applied to a skater jumping straight off the ground as well? Frankly, no. Well, the kinetic energy described before is due to the vertical speed when leaving to floor. When jumping from a flat floor, the speed is wholly from the skaters push. The acceleration in this case is defined by the mass and force.

a = F/m

For a fixed force (push), a greater acceleration is attained with a lower mass. This results in a higher speed leaving the floor. Put simply, lighter objects are easier to accelerate.

Mass is of no consequence once in the air. I was assuming the skater did not jump as such, just was 'stiff' and it is only the ramp which 'pushes' you into the air. This is what I seem to do when on the bigger ramps. (over 45 degrees)