I'd like to suggest that this isn't quite the right question.  At least, it seems to me that modifying the question (in a direction that Theo was hinting) would be more interesting.

The problem with the question as asked is that, for a given category _C_, the mere existence of a faithful functor _C_ --> Set tells you very little indeed.  Perhaps you have some reason for wanting to know that I can't see.  But a condition that seems to have more bite is 'small-concreteness', defined as follows.

Let _C_ be a category.  A set-valued functor _U_: _C_ --> Set is <b>small</b> if it can be expressed as a small colimit of representables.  Call a category _C_ <b>small-concrete</b> if there exists a small, faithful functor _C_ --> Set.  In the special case that _C_ is small, all set-valued functors on _C_ are small and small-concrete = concrete.

It's not too hard to show that a category is small-concrete if and only if it admits a generating set.  (A <b>generating set</b> in a category _C_ is a [small] set _S_ of objects such that, for any distinct maps <i>f, g: a --> b</i> in _C_, there exist _s_ in S and <i>q: s --> a</i> such that <i>fq \neq gq</i>.)  The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see <i>Categories for the Working Mathematician</i>.

You can exploit this as follows.  Suppose you want to show that the category of affine schemes is not small-concrete (which would imply that the category of all schemes isn't either).  Assuming for a contradiction that it <i>is</i> small-concrete, the category Ring of commutative rings has a cogenerating set.  Since Ring is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from Ring to a locally small category has a left adjoint.  I guess it's possible to cook up (or look up) an example of a limit-preserving functor out of Ring that <i>doesn't</i> have a left adjoint.  That would produce the desired contradiction.