What bjorne said is right but incomplete.
The effect is not due to the flux.
Due to Lorentz force electrons which are moving at a distance $r$ from the center with tangential velocity $v=r\omega $ (due to rotation of the disk) experience a radial force (do the vector product to see this) $$F_L=er\omega B $$ where B is the field and $e $ the electrons charge.
This leads to electrons moving towards the periphery and creates a charge discontinuity (also because they leave their nuceli behind) which generates a radial electric field (it is radial because it is due to a radial gradient of charge) $E(r)={dV\over dr} $ and thus a voltage $V (r) $. The fact that $E={dV\over dr}$ is due to the symmetry of the problem: all other components of the gradient vanish as the field only variates in the radial component.
In the steady state the electric ($F_E=eE=e{dV\over dr} $) and Lorentz forces cancel out as nothing should move at equilibrium.
So $$er\omega B=e{dV\over dr} $$
and this gives us an equation for the voltage which when integrated leads to:
$$V(r)=B\omega {r^2 \over 2} $$ with respect to the center of the disk where $V=0$.
Between the edge of the disk the voltage is
$$V (R)=B\omega { R^2 \over 2}$$
which is the same formula you propose.
EDIT: Notice that in this answer I have been neglecting centrifugal force $F_C=m_e\omega^2 r$ acting on the electrons. We can add it to show that the effect, the fact that a voltage arises, is not a purely magnetical effect.
Once again the trick is saying that in the steady state nothing moves, so the total force must be zero. That means:
$$e{dV\over dr} =er\omega B+m_e\omega^2 r$$
or rather
$${dV\over dr} =\left(B+{m_e\over e}\omega\right)\omega r$$
which when integrated leads to
$$V(r)=\left(B+{m_e\over e}\omega\right)\omega {r^2\over 2}$$
Thus even if $B=0$ we still get a voltage $V(R)={m_e\over e}\omega^2{R^2\over 2}$ across the disk.
Yet notice that the effect is small as for an electron ${m_e\over e}\sim 10^{-11}$ (which by the way is in principle measurable from this device).