The following shows that (A) is in some sense trivially true.
Suppose $\mathbb{C}$ and $\mathbb{D}$ are categories and that $(\alpha_X : A_X \to B_X)_{X \in \mathbb{C}}$ is a family of morphisms in $\mathbb{D}$. I hope you would agree that this situation is even more general than the one you describe (since after all one can take $\mathbb{C}$ to be set considered as a category). If $\mathbb{S}$ is the subcategory of $\mathbb{C}$ whose objects are the same as $\mathbb{C}$ and whose morphisms are only identity morphisms, then there are two functors $F, G : \mathbb{S} \to \mathbb{D}$ and a natural transformation such that $F(X) = A_X$, $G(X)=B_X$ and $\theta_X = \alpha_X$.

On the other hand the following is sort of a "counter example" to (A):
Let $\mathbb{C}$ be the category of groups to each group $G$ we can associate two groups $G/Z(G)$ and $\text{Inn}(G)$ the quotient of $G$ by its center and the group of inner automorphisms, respectively. It is well-known that these two groups are isomorphic. Both these assigments seem to be natural however there doesn't seem to be any natural way of making them into functors $\mathbb{C}$ to $\mathbb{C}$. In particular the natural way one might try to make $G/Z(G)$ into a functor would be so that the canonical morphism $G\to G/Z(G)$ is a natural transformation. This is impossible since it would imply that the center can be made into a functor (see [center is not a functor][1]). If we restrict the domain category to the category of groups and surjective group homorphisms, then this passage works. Let us consider now the other assignment (i.e. $G$ maps to $\text{Inn}(G)$). Suppose that $f:G\to H$ is a group homorphism and $\theta : G\to G$ is an inner automorphism (i.e. there exists $g$ in $G$ such that for all $x$ in $G$, $\theta(x)=gxg^{-1}$). One might be tempted to try and define $\text{Inn}(f)(\theta)$ to be the inner automorphism defined by $\text{Inn}(f)(\theta)(y) = f(g)yf(g)^{-1}$. However this does not always determine a group homorphism (i.e. is not always "well defined") unless $f$ is surjective.

I hope that from the above you see that your question is perhaps more about when it is that some construction determines a functor in the expected way. This seems to be a difficult question.

  [1]: http://math.stackexchange.com/questions/158438/why-is-there-no-functor-mathsfgroup-to-mathsfabgroup-sending-groups-to-th