Removability with respect to homeomorphisms is different from the removability with respect to bounded functions mentioned in another answer.
In particular, it is not necessary to have Hausdorff dimension at most 1. Indeed, any quasicircle is removable with respect to homeomorphisms, as mentioned by Hrant.
For much more complicated sets, see Jeremy Kahn's thesis "Holomorphic Removability of Julia sets": He shows that many Julia sets of quadratic polynomials are in fact removable.
http://arxiv.org/abs/math/9812164
(As above, we consider the set in question to be compact.)
In particular, he discusses the notion of "absolute area zero": A set $K$ has absolute area zero if there is no conformal isomorphism from the complement of $K$ to the complement of some set with positive area. Any such set is removable, and any Cantor set that is well-surrounded has absolute area zero.
On the other hand, as has been noted elsewhere, there are many examples of sets that are not holomorphically removable. The simplest example of a Cantor set would be a Cantor set of positive measure. More interesting examples are provided by Chris Bishop, as cited in Misha's answer.
EDIT: You may also wish to look at the paper "Removability theorems for Sobolev functions and quasiconformal maps" by Peter Jones and Stas Smirnov, which contains a number of sufficient conditions for conformal removability: http://www.unige.ch/~smirnov/papers/hr-j.pdf
Graczyk and Smirnov use these criteria to prove removability of a large class of Julia sets.
