Let $R$ be a a commutative ring with a unit element, then one can associate to $R$ a Boolean ring $B(R)$, as in this text by Bergman, last line of page 594. (I guess this is a very classical thing. Explicitly, $B(R)$ is the set of idempotent elements of $R$ with the operations $e \oplus f = e + f - 2ef$ and multiplication inherited from $R$.) Bergman shows that the functor $R \mapsto B(R)$ has a left adjoint - the one you would expect. I suspect that it does not have a right adjoint, but I don't see how to prove this (if it is true).
All ideas would be welcome!
I will make use of Hurkyl's comment above. If this functor has a right adjoint, then it must preserve colimits, and in particular, it must send coproducts to coproducts. This is not the case, so the functor does not have a right adjoint.
The coproduct in the category of commutative rings is well-known to be the tensor product $-\otimes_{\mathbb{Z}}-$, which I will henceforth write as $-\otimes-$. This has the funny property that the coproduct of two nonzero rings may be zero. For a famous example, if $R = \mathbb{Z}/(2)$ and $S = \mathbb{Z}/(3)$, then $R \otimes S = 0$. Thus $B(R \otimes S)$ is the trivial Boolean ring, in which $0 = 1$.
On the other hand, because $R$ and $S$ are fields it follows that $B(R)$ and $B(S)$ are both (isomorphic to) $\mathbb{Z}/(2)$. Because $\mathbb{Z}/(2)$ is the initial object in the category of Boolean rings, the coproduct of $B(R)$ and $B(S)$ in this category readily seen to be $\mathbb{Z}/(2)$ once again.
So the functor $B$ applied to the coproduct of these rings $R$ and $S$ is not isomorphic to the coproduct of the Boolean rings $B(R)$ and $B(S)$.