All the following is simply a simple application of math. If I'm wrong anywhere (and I'm bound to be), please feel free to point it out to me or just edit the post, if you like.
The largest planetoid is Ceres, at almost 1000 km in diameter. If we assume that your planetoid was a bit larger than this, and that Ceres was the asteroid you're trying to slingshot around, we can do some maths and estimate a limit to see if it is truly possible to slingshot the rocket back into the spacecraft, using the largest gravity fields that we can.
At a mass of 94010^18 kg, your asteroid will exert a gravitational force on your (perhaps) 100kg missile that is F = (6.27262 10^12)/$r$^2. This will be for any value of $r$.
If you slingshot a body by a planet, the speed of the missile will then be twice the speed of the planet plus the missile's initial speed. So you should probably be orbiting against the revolution of the asteroid belt. Anyway, the final speed of the missile will be $2U + v$, where $U$ is the speed of the planet and $v$ is the speed of the missile. If we let this quantity be $w$, then you only have 2 variables now determining if you can slingshot back to the spacecraft: $w$ and $r$.
The critical condition here is that given a w (determining your kinetic energy) and an r (determining your potential energy), can we reach an energy quantity that:
a) enters the orbit of the asteroid such that it is captured in the orbit
b) escapes the gravity of the planetoid so the missile does not actually end up orbiting the asteroid
c) gains enough speed that it can slingshot back to the spacecraft in a few minutes.
The escape speed of the asteroid is:
$$v_{escape} = \sqrt{{\frac{2GM}{r}}}$$
So entering the orbit, your $v < v_{escape}$. But since you are exiting the planet, your $w > v_{escape}$.
Since $2U + v = w$, then we want everything expressed in terms of $v$:
So $$v < v_{escape} < w $$
$$v < v_{escape} < 2U + v $$
We're going to assume that the missile will be whizzing just above the surface of the planetoid (for maximum energy), so $v_{escape} = 0.51$ km/s and $r = 490$ km. 921
Plus, we'll assume the missile is going at the fastest speed of satellite we have ever produced, which is $v = 70$ km/s. Substituting:
$$70 < 0.51 < 2U + 70 $$
So, this can't happen. Your missile should be going slower than this. What we need is a speed that is less than 0.51 km/s when we enter, so let's go for 0.50 km/s.
$$0.50 < 0.51 < 2U + 0.50 $$
To determine $U$, we need to estimate the distance your missile is going to be travelling. If we let $x$ be the distance of the spacecraft from the planetoid, and the planetoid is not moving, then if the missile is whizzing just above the surface of the planetoid, your missile will be travelling a distance of $2x + \pi r$. The missile will be travelling at different speeds throughout this course, so let's sum up the times required. Let's let the acceleration of the missile experiences be $a$.
The acceleration is the square root of the sum of the squares if the tangential and radial acceleration.
$$a = \sqrt{a_t^2 + a_r^2}$$
We know the acceleration at the surface of Ceres, which is our $a_r$. We also know $a_t$ because that's just the acceleration to get from $v$ to $w$.
$$a = \sqrt{(\frac{w - v}{t})^2 + (0.028g)^2}$$
Playing around with this equation, we will get
$$t = \frac{w - v}{\sqrt{a^2 - a_r^2}}$$
And this is the time it will take the missile to whiz by half the surface of Ceres. 234