There is a more general result that this follows from. This is in Gordon's paper cited above; see Theorem 4. In any event, you don't really need the handle arguments in Marco's answer.

Here is a sketch of Gordon's construction, in slightly different language. Of $K$ and $K'$ are concordant, then for any $p/q$ there is a homology cobordism between $S^3_{p/q}(K)$ and $S^3_{p/q}(K')$ with fundamental group normally generated by the meridian of $K$ or the meridian of $K'$.

To see this, consider a tubular neighborhood $S^1 \times D^2 \times I$ of the concordance, and do $p/q$ surgery at each level. The meridian of the concordance is conjugate to the meridian of either end, and normally generates the fundamental group of the complement, and hence the fundamental group of the homology cobordism.

When $K$ is slice, apply this to $K' = $ the unknot to get a homology cobordism to a lens space $L(p,q)$ with fundamental group $\mathbb{Z}/p$. In particular when $p=1$, you get a simply connected homology cobordism to $S^3$, which gives a contractible $4$-manifold with boundary $S^3_{1/q}(K)$ as desired.