Let us suppose that we have a ribbon embedding $S^n \rightarrow S^{n+2}$ for $n\geq 3$. Call this knot $K$. By a theorem of Levine (and Trotter for $n=3$ I believe) we know $K$ is unknotted if the complement is homotopy equivalent to $S^1$. The question which I wonder is, does it suffice to show that the fundamental group is $\mathbb{Z}$, given the fact that $K$ is ribbon? For $n=2$ this would be true, as Freedman mentions (that having infinite cyclic fundamental group implies being homotopy equivalent to a circle) but I am not terribly familiar with the highdimensional case.

If the fundamental group of the complement is the integers there's still the various Alexander modules for the complement. Ribbon knots aren't immune from having nontrivial Alexander modules in dimensions higher than 1. Take a look at Kearton's papers on "simple nknots" 


I found a paper which discusses this issue: Ribbon knots and ribbon disks from Asano, Marumoto, and Yanagawa. They establish that for $n\geq 3$ a ribbon knot with infinite cyclic fundamental group is trivial. http://ir.library.osakau.ac.jp/metadb/up/LIBOJMK01/ojm18_01_12.pdf 

