One can prove that for any non-zero ring $R$ the category $R$-Mod$^{op}$ is not a category of modules. Indeed any category of modules is Grothendieck abelian i.e., has exact filtered colimits and a generator. So for $R$-Mod$^{op}$ to be a module category $R$-Mod would also need exact (co)filtered limits and a cogenerator. It turns out that any such category consists of just a single object.
I believe this is stated somewhere in Freyd's book Abelian Categories but I am not sure exactly where off the top of my head. Edit it is page 116.
Further edit: For categories of finitely generated modules here is something else, although it is more in the direction of the title of your question than what is in the actual body of the question. Suppose we let $R$ be a commutative noetherian regular ring with unit and let $R$-mod be the category of finitely generated $R$-modules. Then we can get a description of $R$-mod$^{op}$ using duality in the derived category.
Since $R$ is regular every object of $D^b(R) \colon= D^b(R-mod)$ is compact in the full derived category. The point is that
$$RHom(-,R)\colon D^b(R)^{op} \to D^b(R) $$
is an equivalence (usually this is only true for perfect complexes, but here by assumption everything is perfect). So one can look at the image of the standard t-structure (which basically just "filters" complexes by cohomology) under this duality. The heart of the standard t-structure is $R$-mod sitting inside $D^b(R)$ so taking duals gives an equivalence of $R$-mod$^{op}$ with the heart of the t-structure obtained by applying $RHom(-,R)$ to the standard t-structure.
In the case of $R =k$ a field then this just pointwise dualizes complexes so we see that it restricts to the equivalence $k$-mod$^{op}\to k$-mod given by the usual duality on finite dimensional vector spaces.
As another example consider $\mathbb{Z} $-mod sitting inside of $D^b(\mathbb{Z})$ as the heart of the standard t-structure given by the pair of subcategories of $D^b(\mathbb{Z})$
$$\tau^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i>0\}$$
$$\tau^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i\leq 0\} $$
It is pretty easy to check that $RHom(\Sigma^i \mathbb{Z}^n, \mathbb{Z}) \cong \Sigma^{-i} \mathbb{Z}^n$ and $RHom(\Sigma^i \mathbb{Z}/p^n\mathbb{Z}, \mathbb{Z}) \cong \Sigma^{-i-1}\mathbb{Z}/p^n\mathbb{Z}$ so that this t-structure gets sent to
$$\sigma^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i> 0, \; H^{0}(X) \; \text{torsion}\}$$
$$\sigma^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i< 0, H^0(X) \; \text{torsion free}\} $$
using the fact that objects of $D^b(\mathbb{Z})$ are isomorphic to the sums of their cohomology groups appropriately shifted.
So taking the heart we see that
$$\mathbb{Z}-mod^{op} \cong \sigma^{\leq 0} \cap \Sigma\sigma^{\geq 1} = \{X \; \vert \; H^{-1}(X) \; \text{torsion free}, H^0(X) \; \text{torsion}, H^i(X) = 0 \; \text{otherwise}\} $$
with the abelian category structure on the right coming from viewing it as a full subcategory of $D^b(\mathbb{Z})$ with short exact sequences coming from triangles. It is the tilt of $\mathbb{Z}$-mod by the standard torsion theory which expresses every finitely generated abelian group as a torsion and torsion free part.
Concerning your first question I have a couple of suggestions: first coisotropic is in some sense the best we can have in a truely Poisson situation: there is nothing like lagrangian (unfortunately).
As you already said, lagrangian sometimes is associated to pure states in the quantum regime, here the main argument is coming from the WKB approximation in physics, which has some very interesting mathematical formulations: in the booklet of Bates&Weinstein (you probably know) you can find a lot of these ideas.
However, I would like to draw your attention to some other point: Having a coisotropic $I$ in some Poisson algebra $A$ (so we left geometry, purely algebraic setting) then the thing you can do classically is reduction: You take the idealizer $B \subseteq A$ of $I$ with respect to the Poisson bracket and this turns out to be the largest Poisson subalgebra of $A$ having $I$ as Poisson ideal. So the quotient $B/I$ is again a Poisson algebra. In your favorite geometric setting with nice assumption this corresponds precisely to the Poisson algebra of functions of the (Marsden-Weinstein) quotient: But we see something more:
$A/I$ is a $A$-left module (sure) and it becomes also a $B/I$ right module (just check that things are well-defined). In fact, a little exercise shows that if $A$ has a unit (let's assume that) then $B/I = End_A(A/I)^{op}$. So we are in some sense even very close to a Morita context (it is not, though...)
Usually I don't like to do that: but to make a little advertisement for some own work, I have a quite detailed paper with Simone Gutt on the above reduction proceedure where we investigate the relations of the representation theories of the big algebra $A$ and the reduced algebra $B/I$ :)
Now the remarkable thing is that this (still classical) bimodule structure might have good chances to survive quantization. In fact (and here one should quote Martin Bordemann's long french preprint as well as the works of Cattaneo&Felder) under certain geometric conditions deformation quantization gives indeed such a quantization.
So the noncommutative version is a left ideal $I$ and then the above quotient proceedure just works the same on the algebraic level. Of course, in DQ the trickey question is whether $B/I$ is still something like the quantized functions on the classical Marsden-Weinstein quotient and even more trickey: whether the classical coisotropic ideal can be quantized into a left ideal at all. For this there are obstructions, even local ones, in the Poisson setting, while it works locally in the symplectic setting by taking an adapted Darboux chart. Globally, it is also trickey in the symplectic setting: Martin Bordemann discusses this in detail...
OK: the conclusion is something like coisotropic ideals are used for reduction and their quantization will be left ideals used in the same way. Both lead to the above bimodule structures on $A/I$ which is geometrically the coisotropic submanifold itself.
As a small warning: it is not true in deformation quantization that all modules (of interest) arise this way. There are other modules which have their support say on points: one can use $\delta$-functionals as positive functional (after some correction terms) and get a GNS like construction also in formal deformation quantization. Then in this case, the module has sort of support on that point...
Ah, the second question: never thought about that in detail, but perhaps the above picture gives some ideas on "reverse engineering"..?
Best Answer
Here is one precise statement of how quantization is not a functor:
5) There is no functor from the classical category $\mathcal C$ of Poisson manifolds and Poisson maps to the quantum category $\mathcal Q$ of Hilbert spaces and unitary operators that is consistent with the cotangent bundle/$\frac12$-density relation (explained below).
The result is due to Van Hove, in "Sur le probleme des relations entre les transformations unitaires de la mecanique quantique et les transformations canoniques de la mecaniques classique." This is an old paper and I can't find a link for it, but the reference I found it in is Weinstein's "Lectures on Symplectic Manifolds."
By "cotangent bundle/$\frac12$-density relation" I mean the following: if $\mathcal M$ is the category of smooth manifolds and diffeomorphisms, we have a cotangent functor $\mathcal M \to \mathcal C$. This assigns to each manifold its cotangent bundle with the canonical symplectic structure, and to each diffeomorphism the induced symplectomorphism of cotangent bundles.
We also have a natural functor $\mathcal M \to \mathcal Q$. For any smooth manifold $X$ consider the bundle of complex $\frac12$-densities on $X$. (What is the bundle of complex $s$-densities? Well, the fiber over a point $x \in X$ is the set of functions $\delta_x: \bigwedge^{top} T_xX \to \mathbb{C}$ such that $\delta(cv) = |c|^{s}\delta(v)$.) If $\delta^1$ and $\delta^2$ are smooth compactly-supported $\frac12$-densities, their pointwise product $\delta^1 \bar{\delta^2}$ is a compactly supported 1-density which we can integrate to get a complex number. This turns the space of all such sections into a pre-Hilbert space, the completion of which is what our functor assigns to the manifold $X$. As we would hope for, the canonical nature of the construction lets us assign unitary operators between Hilbert spaces to diffeomorphisms between smooth manifolds, hence is functorial.
(Note: If we choose a volume form on $X$, the above procedure produces something isomorphic with the space of $L^2$ functions on $X$ with respect to this form, but to get something functorial we want a canonical construction.)
From this pair of functors $\mathcal M \to \mathcal C$ and $\mathcal M \to \mathcal Q$ we get a product functor $\mathcal M \to \mathcal{C} \times \mathcal{Q}$. The image of this functor is a subcategory of $\mathcal C \times \mathcal Q$ which we will call the "cotangent bundle/$\frac12$-density relation." (The word relation is meant in the same sense that an ordinary relation between two sets is a subset of their product).
Now we can clarify just what is meant by our original statement: there is no functor $\mathcal C \to \mathcal Q$ whose graph contains the cotangent bundle/$\frac12$-density relation. The reasons why this is a desirable condition come from physics and are beyond me, but roughly speaking I think the point is that there exists a good idea of what a quantization functor is supposed to do to cotangent bundles.