Do an unlinked trefoil and figure-eight cobound an annulus in $B^4$? Let $K_1$ the trefoil (left or right hopefully does not matter?) and let $K_2$ be the figure-eight knot in $S^3 = \partial B^4$.  Are there any smooth properly embedded annulus $A$ in $B^4$ with $\partial A = K_1 \coprod K_2$?
I'm really interested in knowing some general obstructions for this sort of thing for more general knots $K_1,K_2$ so any ideas are welcome.  I figured that I would ask the question with two specific knots just for fun.   
 A: The relationship you're asking for is called concordance.  Determining if knots are concordant is quite difficult: there are many concordance invariants, but no kind of global picture of what it means.
One concordance invariant is the knot signature.  The trefoil has signature $-2$ and the figure-8 knot has signature $0$, so they are not concordant.  I got this data from the wonderful website Table of Knot Invariants.
As far as where to read about knot concordance, a good place to start is Livingston and Naik's book-in-progress here.

EDIT: Knots $K_0$ and $K_1$ are usually said to be (smoothly) concordant if there is a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$.  In the comments, I was asked to sketch why this is the same as having an annulus connecting $K_0$ and $K_1$ in $B^4$ (assuming that $K_0$ and $K_1$ are unlinked).
There are a lot of ways to prove this.  Here's a brief sketch of one of the easiest.  If $D$ is a small open round ball in $S^3$, then
$$\left(S^3 \times [0,1]\right) \setminus \left(D \times [0,1]\right) \cong B^4.$$
There is a tiny issue in that the right hand side is a manifold with corners, but as always corners can be smoothed.  Using this diffeomorphism, we immediately see that if there is a smoothly embedded annulus in $B^4$ connecting $K_0$ and $K_1$, then after possibly moving $K_0$ and $K_1$ around we can obtain a smoothly embedded annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$; this annulus avoids $D \times [0,1]$.  Conversely, if we can find an annulus in $S^3 \times [0,1]$ connecting $K_0 \times \{0\}$ and $K_1 \times \{1\}$, then homotoping everything we can assume that this annulus avoids some $D \times [0,1]$, and then transport it to an annulus in $B^4$ connecting $K_0$ and $K_1$.
The above sketch elides various identification we are making, but once you understand the geometry (draw a picture!) you'll see that they are all pretty clear.
