Given a commutative ring $R$, finitely generated projective $R$-modules $M$ are exactly those for which tensoring with $M$ has not just a right adjoint, but also a left adjoint (or equivalently, commutes with all limits (not just finite ones, as is the case when $M$ is merely flat)).
This suggests the following:
Question: Is $C^{op}$ cartesian closed, where $C$ is the category of commutative $R$-algebras whose underlying $R$-module is finitely generated projective?
Note that the tensor product of commutative $R$-algebras is their coproduct in the category of commutative(!) $R$-algebras. Also, $R$ itself is a finitely generated projective $R$-module, while the tensor product of two finitely generated projective $R$-modules is easily seen to itself be finitely generated projective. Hence, the coproduct restricts to the full subcategory $C$, which is then the product in $C^{op}$.
Perhaps, using cocommutative $R$-coalgebras might help.
[I will use the language of schemes, for instance identifying $C^{op}$ with a certain subcategory of $R$-schemes.]
No. For instance, let $R=k$ be an infinite field and consider the object $T=\operatorname{Spec} k[x]/(x^2)$ of $C^{op}$. If an exponential object $T^T$ existed, then maps $\operatorname{Spec}k\to T^T$ would be in bijection with maps $T\cong \operatorname{Spec} k\times T\to T$. But there are infinitely many maps $T\to T$ (one for each element of the field) and only finitely many maps $\operatorname{Spec} k\to X$ for any object $X$ of $C^{op}$ (since each such map is uniquely determined by a point of $X$ and $X$ is finite since it is Spec of an artinian ring).
Note though every object of $C^{op}$ is exponentiable in the full category $\mathrm{Aff}_R$ of affine $R$-schemes (i.e., the opposite category of commutative $R$-algebras). This follows easily from the adjoint functor theorem. Note moreover that if $X=\operatorname{Spec} A$ is an object of $C^{op}$ then the product functor $-\times X:\mathrm{Aff}_R\to \mathrm{Aff}_R$ can be factored as a composition $\mathrm{Aff}_R\to \mathrm{Aff}_A\to \mathrm{Aff}_R$ where the first functor is the base change functor and the second functor is the forgetful functor. Each of these functors has a right adjoint: the right adjoint to base change is known as Weil restriction and the right adjoint to the forgetful functor is the base change functor. Thus the exponentiation $Y^X$ can be identified with $\operatorname{Res}_{A/R}(Y\times_R X)$ where $\operatorname{Res}_{A/R}$ is Weil restriction from $A$ to $R$.