Following up on Akhil's plan, a good source of projective modules which are not free is via nontrivial vector bundles: smooth sections of the bundle form a projective module over the ring R of smooth functions, but a non-free module if the bundle is nontrivial.

So, you could do something simple like this: apply the exterior algebra construction to the cotangent bundle over S^2; this gives a nonfree projective R-module N of the sort Akhil referred to. Then the normal bundle to N gives a nonfree projective P, and if 1 denotes the trivial line bundle, the direct sum 1 + P gives a commutative R-algebra M in a mindless way by defining all products of sections of P to be zero. The product M x N is then free, but the center M is not.

Following up on Akhil's plan, a good source of projective modules which are not free is via nontrivial vector bundles: smooth sections of the bundle form a projective module over the ring R of smooth functions, but a non-free module if the bundle is nontrivial.

So, you could do something simple-minded like this: let P be the module of sections of the tangent bundle of RP^2; put a commutative algebra structure on M = R + P where all products of elements of P are zero. As an R-module, M is not free. Let N be the module of sections of the exterior algebra bundle of the normal bundle to an embedding of RP^2 in R^4. The Whitney sum of the tangent bundle and the normal bundle is trivial; therefore the algebra M x N forms a free R-module, whose center M is not free.