Buoyancy a consideration?
From YOUR link-
The terminal velocity of a falling object is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.[
and latter-
Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass m has to be reduced by the displaced fluid mass \rho\mathcal{V}, with \mathcal{V} the volume of the object. So instead of m use the reduced mass m_r=m-\rho\mathcal{V} in this and subsequent formulas.
So, what you've done is use on equation that doesn't take all aspects of the situation into account.
Applying THAT principle will show that the less dense object will accelerate slower at the outset, so the two objects will NEVER fall at the same rate.