The escape velocity $v_e$ required to escape a body of mass $M$ from a distance $r$ is equal to
$$v_e=\sqrt{\frac{2GM}{r}}$$
And it happens that the escape velocity is the same in all directions. Regardless of whether your trajectory is directly away from the planet or oblique, a velocity of $v_e$ will send you off into space.
The mass of Saturn is about $M\approx 5.68\cdot 10^{26}\space\text{kg}$, and the gravitational constant $G$ is about $G\approx 6.67\cdot 10^{-11}\space\text{Nm}^2/\text{kg}^2$. The radius of Saturn’s Phoebe ring (which is the farthest out of all its rings, and therefore the easiest to escape from) is about $r\approx 1.75\cdot 10^{10}\space\text{m}$. This means that escape velocity from the ring system is about
$$v_e \approx \sqrt{\frac{(2)(6.67\cdot 10^{-11})(5.68\cdot 10^{26})}{1.75\cdot 10^{10}}}\approx 2080.9\space\text{m/s}$$
The fastest land animal on Earth - the cheetah - can travel at a speed of $33.53 \space\text{m/s}$, which is much slower than the required velocity. And it doesn’t sound like your space-hoppers are anywhere near as fit as a cheetah.
However, there is another potential problem. Since Saturn’s rings are also moving at a constant velocity, the space-hoppers could possibly piggyback on the rings’ velocity to reach escape velocity.
The orbital speed of an object orbiting at a distance $r$ is approximately
$$v_o = \sqrt{\frac{GM}{r}} = \frac{v_e}{\sqrt{2}}$$
So we have that Saturn’s rings are moving at an orbital velocity of about
$$v_o\approx 1471.4\space\text{m/s}$$
This means that the extra velocity that space-hoppers must generate in order to escape is given by
$$v_e - v_o \approx 609.5\space\text{m/s}$$
...which is still much faster (more than $18$ times faster) than a cheetah can travel. So it’s pretty unlikely that your space-hoppers can escape, unless they have extremely powerful spring-like appendages.
As for crashing into the planet - that’s always a danger, if the space-hoppers wander too close to the inner edge of the rings. If they propel themselves beyond this inner edge, there will be nothing for them to grab onto (not even an atmosphere to fly in) to halt their acceleration towards the planet. The best bet would be to stay near the outer edge of the rings, since the change of accidentally escaping is practically nil.
EDIT: @Daron has considered the possibility of escaping the ring system but not the planet’s orbit, taking up an orbit outside of the planet’s rings. The calculations above haven’t accounted for that possibility, just the possibility of flying out into space forever. Presumably, if a baby space-hopper flew into a more distant orbit, all hope would not be lost and its parents might devise away of drawing it back. Let’s deal with this situation numerically.
We can calculate the radius of the farthest possible orbit that a space-hopper can fling itself into using a calculation involving the energy of the orbit. The total energy of an orbiting object with mass $m$ at a distance $r$ from a planet of mass $M$ is given by
$$E=-\frac{GMm}{2r}$$
The energy $dE$ required to change the orbit radius by a small amount $dr$ is about
$$dE\approx \frac{GMm}{2r^2}\cdot dr$$
Suppose that your space-hopper is about the mass of a cow, making it about $1.4\cdot 10^6$ grams. Then the approximate energy required to increase the orbit radius by $1$ meter starting from the Phoebe ring is about $0.087\space\text{N}$. That’s not very much energy.
The bite force of an alligator is around $13,000\space\text{N}$, so if your space-hopper can jump as hard as an alligator can chomp, then it could increase the radius of its orbit by around $1.5\cdot 10^{5}\space\text{m}$. Oof.
On second thought, better stay towards the middle of the rings.
