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 $C \to \mathbf{Set}$ 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 $U: C \to \mathbf{Set}$ is **small** if it can be expressed as a small colimit of representables.  Call a category $C$ **small-concrete** if there exists a small, faithful functor $C \to \mathbf{Set}$.  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 **generating set** in a category $C$ is a [small] set $S$ of objects such that, for any distinct maps $f, g: a \to b$ in $C$, there exist $s \in S$ and $q: s \to a$ such that $fq \neq gq$.)  The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see *Categories for the Working Mathematician*.

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 *is* small-concrete, the category $\mathbf{Ring}$ of commutative rings has a cogenerating set.  Since $\mathbf{Ring}$ is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from $\mathbf{Ring}$ 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 $\mathbf{Ring}$ that *doesn't* have a left adjoint.  That would produce the desired contradiction.