You asked about the advantages of links in $S^3$ over links in $\mathbb R^3$ (or $B^3$, which I prefer). Here's an advantage of $B^3$ over $S^3$: Khovanov homology is a (functorial) invariant of links in $B^3$, but so far as I know there is no known proof that it is an invariant of links in $S^3$.

More specifically, the standard literature on Khovanov homology constructs a functor from this category

- Objects: links in $B^3$
- Morphisms: Isotopy classes of isotopies in $B^3\times I$ (or more generally isotopy classes of cobordisms in $B^3\times I\;$).

to the category of bigraded chain complexes and homotopy classes of chain maps. The proof involves defining a chain map for each Reidemeister move and elementary cobordism, and then checking that these generators satisfy various "movie move" relations.

If you want to prove an analogous theorem for links in $S^3$, then there is an additional "global movie move" to check. It corresponds to the case where a 1-parameter family of isotopies in $S^3\times I$ (i.e. a second order isotopy) transversely intersects {south pole of $S^3$} $\times I$. I don't think anyone knows how to prove invariance under this global movie move. Perhaps it's not even true. (Scott Morrison and I announced a proof a couple of years ago, but we later discovered a sign issue. We're currently writing up a $\mathbb Z/2$ version with Chris Douglas. If anyone knows how to prove/disprove the relation with $\mathbb Z$ coefficients, please speak up!)

Morally, the reason for this difficulty is that there is no nice projection of links in $S^3$ to link diagrams in $S^2$. The obvious attempt at a map from $S^3$ to $S^2$ is not well-defined at the north and south pole of $S^3$, and this ambiguity actually matters for categorical/functorial link invariants.

In summary, if you are studying categorical link invariants (like Khovanov homology) in terms of planar projections, you are dealing in links in $B^3$, not $S^3$.