Gravitational time dilation compensated by acceleration

Both clocks have the same proper acceleration. The calculation of the proper acceleration is described in What is the weight equation through general relativity? The proper acceleration for an object stationary at a distance $r$ from the centre of the Earth turns out to be:

$$ A = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$

It makes no difference that one clock is stationary on a mountain at the distance $r$ from the centre of the Earth while the other is stationary in a helicopter at the distance $r$ from the centre of the Earth.

The reference frames of clock $A$ and clock $B$ are equivalent. They are stationary with respect to the spacetime given by the earth mass/energy at the same radial coordinate. They measure a proper acceleration as the frames are not along a geodesic.

The time of clock $A$ and clock $B$ runs with the same rate.