Sure, that'd just be saying that inertial mass is different than gravitational mass
$m_i a = m_g g$
So that masses wouldn't be the same, they wouldn't cancel out anymore, and $a$ would be no longer $g$, but
This would give you the acceleration. Now you'd have to integrate this acceleration to get velocity and position.
If $g$ is constant it would just be the typical Uniformly Varied Straight Movement.
$y_F=y_0 + v_ot - at^2/2$
But $a$ wouldn't be $g$ now, but the upper expression.
as I say in my comment, I'm supposing that, asthe equivalence principle says that an accelerated observer is indistinguishable from being in a gravitational field, and that's because $m_g=m_i$; if they aren't the same anymore, each object will fall at a different rate, a rate given by their own ratio $m_g/m_i$. I am assuming that $m_g$ is not the same as $m_i$ now, nor it is the same ratio for all bodies. If the ratio were the same, we would still have $a\propto g$, so we would only need to redefine $G$ so that the gravitywere actually the same as $a$.
However, if the equivalence principle doesn't hold, each body should have a different ratio, and so they'd fall with different accelerations.