Note: Being no expert in any area of topology, I might make some major errors in describing what I'm envisioning here.
Imagine three-dimensional Cartesian space with any point $P$ in it represented as $(x,y,z)$. Now let there be a differentiable function $f(x,y)$ such that $f(x,y) = z$. Let one $z$-value exist for every ordered pair $(x,y)$. Now, let there be two points $P_0(x_0,y_0,z_0)$ and $P_1(x_1,y_1,z_1)$ on the graph of $f$.
On to the physics part.
Now that we have established $f$, let a three-dimensional frictionless surface be modeled by $f$, with the force of gravity pulling toward the $-z$ direction. Let there be a sphere of arbitrarily small radius and arbitrary mass tangent to point $P_0$ on the surface that has a velocity vector $<dx_{sphere}/dt,dy_{sphere}/dt>$. For any point $P_1$, can we assign $dx_{sphere}/dt|_{t=0}$ and $dy_{sphere}/dt|_{t=0}$ such that the ball eventually becomes tangent to ("rolls over") $P_1$ after an arbitrary span of time? (Ignore air resistance.)
Given the constraints of your problem, the answer is no. Here is an example:
Let's create a surface that is independent of $y$, so we constrain our problem on the $xz$ curve. Say the particle starts from $(0,0)$. Then make the particle go up a "hill" with a height $h$, with a small slope. Now put a very narrow hill, say 10 times as high, at a short distance. If the sphere would have enough speed to go over the second hill, it would instead bounce back from the side of the hill 2. So any point on the other side of hill 2 are unavailable. Something like this: