In lecture, I was told that the converse of this statement was true, that is, of a vector-valued function that is along a sphere centered at the origin, the acceleration is always opposite the position. For example $(\cos{2x},\sin{2x},0)$ has a second derivative $(-4\cos{2x},-4\sin{2x},0)$, which is opposite, and $(\cos{2x},\sin{2x},0)$ is the equation of a line along a sphere.
What I'm wondering is that, if given a function $(f(x),g(x),h(x))$ with second derivative $(-f(x),-g(x),-h(x))$, can we say for sure that the function is along a sphere or do we need more information?
We need more information, yes. Any movement governed by an ideal spring of unit strength, centered at the origin, and with neutral length $0$ will follow that rule, and that rule describes only such motion. Note that movement on a sphere does not follow that rule once you slow down, speed up, or turn to the side.
On a sphere, it is only when moving at constant speed along a great circle you get acceleration which is the opposite of position. In fact, you also need said speed to match up with the radius of the sphere if you want the length of position and acceleration to be the same. Unit speed on the unit sphere will do, for instance, and you will need to go faster on a larger sphere.