The clock you throw into the air would run faster.
Doing the calculation rigorously is quite involved, but there is a simple agrument we can use to see why the thrown clock records more time that the stationary clock, and we do this by comparing it to the twin paradox.
Suppose twin $A$ is stationary and twin $B$ is in a rocket. At time zero twin $B$ passes twin $A$ at some velocity $v$, and at that moment they synchronise their clocks. Twin $B$ now starts a constant acceleration towards twin $A$. This constant acceleration gradually slows twin $B$, brings them to a halt then accelerates them back towards twin $A$. When they meet again $A$'s clock shows a time $t_A$ and $B$'s clock shows a time $t_B$. The twin paradox is that $t_B \lt t_A$ i.e. the accelerating twin's clock shows less time.
And this is the key point:
the accelerating twin's clock shows less time than the non-accelerating twin's clock
(If you're interested this type of motion is discussed in What is the proper way to explain the twin paradox?)
In your situation we also have a constant acceleration, the gravitational acceleration $g$, and we have two observers. The key point is that it is the observer who is stationary on Earth who is the accelerating one, therefore it is the observer stationary on Earth who records less time i.e. whose clock runs slower.
Thios may seem odd, because after all it seems like the thrown object that is doing the accelerating. However the acceleration that matters here is the proper acceleration. The proper acceleration of you standing on the Earth's surface is non zero while the proper acceleration of the thrown clock is zero (give or take a bit of air resistance).