Free-forgetful adjunction for modules over monoids in the enriched functor category $\mathcal{V}^\mathcal{C}$. Let $(\mathcal{V},\otimes,I)$ be as closed, symmetric monoidal, bicomplete category and $\mathcal{C}$ a skeletally small $\mathcal{V}$-category. Assume that $\mathcal{C}$ itself has a monoidal structure $\oplus:\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$ with unit $\emptyset$.
Then the category $\mathcal{V}^\mathcal{C}$ of $\mathcal{V}$-enriched functors $\mathcal{C}\rightarrow\mathcal{V}$ is itself naturally enriched over $\mathcal{V}$, with hom-objects defined by the enriched ends
$$\mathcal{V}^\mathcal{C}(X,Y)=\int_{c\in\mathcal{C}}\mathcal{V}(X_c,Y_c),\qquad X,Y\in\mathcal{V}^\mathcal{C}.$$
Furthermore, this functor category carries a monoidal structure given by Day convolution. In detail, for $X,Y\in\mathcal{V}^\mathcal{C}$ we define their product $X\widehat\otimes Y$ as the enriched left Kan extension of $X_-\otimes Y_-:\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{V}$ along $\oplus:\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$. To be concrete we can understand it by the coend formula
$$(X\widehat\otimes Y)_c=\int^{d,e}\mathcal{C}(d\oplus e,c)\otimes X_d\otimes Y_e.$$
It is an easy application of the $\mathcal{V}$-Yoneda Lemma that the unit for this product is the corepresented $\mathcal{V}$-functor $\mathcal{C}(\emptyset,-)$.
Now by a monoid in $\mathcal{V}^\mathcal{C}$ we understand a monoid for the convolution product. Specifically one such is a $\mathcal{V}$-functor $R$ together with $\mathcal{V}$-enriched product $\mu:R\widehat\otimes R\Rightarrow R$ and unit $\eta:\mathcal{C}(\emptyset,-)\Rightarrow R$ natural transormations  that make the expected diagrams commute. By a module $M$ for a monoid $R$ we again mean with respect to the convolution product. That is, a $\mathcal{V}$-functor $M$ together with a $\mathcal{V}$-natural transformation $\rho:M\widehat\otimes R\Rightarrow M$ that makes the obvious associativity and unit diagrams commute.
Now if $M,N$ are $R$-modules in $\mathcal{V}^\mathcal{C}$, then we can define a map of $R$-modules between them to be a $\mathcal{V}$-natural transformation $f:M\Rightarrow N$ which makes the obvious diagrams commute. In this way we get a subcategory $R\mbox-mod_\mathcal{C}$ of $R$-modules in $\mathcal{V}^\mathcal{C}$, which we can enrich with hom-objects defined by means of equalisers
$$R\mbox-mod_\mathcal{C}(M,N)\dashrightarrow\mathcal{V}^\mathcal{C}(M,N)\rightrightarrows\mathcal{V}^\mathcal{C}(M\widehat\otimes R,N)$$
where the top arrow is $\rho_M^*$ and the bottom arrow is $\rho_{N*}(-\widehat\otimes R)$.
Now, onwards to the actual question. Let $X\in\mathcal{V}^\mathcal{C}$ be any functor and $R$ a monoid in $\mathcal{V}^\mathcal{C}$. Then $X\widehat\otimes R$ is the 'free' module on $X$, and for any $R$-module $M$ we expect a free-forgetful natural isomorphism
$$R\mbox-mod_\mathcal{C}(X\widehat\otimes R,M)\cong\mathcal{V}^\mathcal{C}(X,M).\qquad (\ast)$$
This is exactly what happens in the classical case, in which the theory reduces to that of rings and modules. However, I could not myself work through the details, although they are probably/hopefully not too complicated.

Can someone fill in the details for why the previous adjunction $(\ast)$ should/should not hold?

Also, if I have made any mistakes above recalling the details, please do point it out.
 A: $\newcommand{\rotimes}{\hat\otimes}$
Here's a sketch of a proof, I suspect it should be enough : 
Step 1 : Prove the result in an un-enriched case, that is : if $(\mathcal V, \otimes, I)$ is a monoidal category, $R$ a monoid-object in $\mathcal V$, then $X\mapsto X\otimes R$ is left-adjoint to the forgetful functor $(R-mod)^0 \to \mathcal V$ (where $(R-mod)^0$ is the unenriched category of $R$-modules). This involves some reasonable diagram-chasing, if you want more details I can add them but it's not too hard : define the obvious adjunction maps and draw the correct diagrams to see that they are inverse. 
Step 2: Use the Yoneda lemma. More precisely, let $(\mathcal V,\otimes, I)$ be a closed symmetric bicomplete monoidal category, $(\mathcal M, \rotimes, e)$ a $\mathcal V$-enriched monoidal category ($\rotimes$ is an enriched bifunctor - though to be fair I'm not sure it needs to be enriched) which is tensored over $\mathcal V$, with tensor $\odot$. 
I'll use the following notations : $\hom$ is the hom-set in $\mathcal V$ and in $\mathcal M$, $\newcommand{\rhom}{\mathcal{Hom}}$ $\rhom$ is the hom-$\mathcal{V}$-object in $\mathcal M$ (note that we have, by definition of hom-set for an enriched category : $\hom(X,Y) = \hom(I, \rhom(M,N))$ for $M,N\in \mathcal M$). Then the tensoring $\odot$ should satisfy $\hom(Y, \rhom(M,N)) \cong \hom(Y\odot M,N)$ (you can probably even ask for an enriched adjunction here, but it's not necessary, so I'm not getting into that)
Tensoring should also be compatible with $\rotimes$ in the sense that $Y\odot (M\rotimes N) \cong (Y\odot M)\rotimes N$ and this isomorphism should be compatible with all the data in $\mathcal M$: associator, unitor. 
With all this under the belt we can see : 
For any monoid-object $R$ in $\mathcal M$ and any $R$-module $M$, $Y\odot M$ is canonically an $R$-module with the obvious structure maps. Then we have the following string of isomorphisms $$\hom(Y, \rhom(X,M)) \cong \hom(Y\odot X, M) \cong \hom_{(R-mod)^0}(Y\odot X\rotimes R, M) \cong \hom(Y, R-mod_{\mathcal M}(X\rotimes R, M))$$
where $R-mod_{\mathcal M}$ is the enriched category of $R$-modules as you have defined it (the definition makes sense for any $\mathcal M$ with the properties I gave), and where the isomorphisms are justified as follows : 
the first one is just the $\odot \dashv \rhom$ adjunction; the second one is from step 1 (since we're dealing with honest modules), and the third one is to justify : 
Note that a morphism $Y\to R-mod_{\mathcal M}(M,N)$ ($M,N$ arbitrary $R$-modules) is the same as a morphism $Y\to \rhom(M,N)$ equalizing the two arrows $\rhom(M,N)\rightrightarrows \rhom(M\rotimes R, N)$. By $\odot \dashv \rhom$ adjunction, this is the same as a map $Y\odot M \to N$ making the following diagram commute : 
$\require{AMScd} \begin{CD}Y\odot M @>>> N \\
@AAA @AAA \\
Y\odot M \rotimes R @>>> N\rotimes R\end{CD}$
which is precisely a module morphism $Y\odot M\to N$; and then we apply this to $X\rotimes R$ and $M$ instead of $M,N$.
It follows (from the unenriched Yoneda lemma) that we have the desired enriched adjunction.
We then check that $\mathcal{V^C}$ satisfies the correct hypothesis, I did that bit in my head so I'm not 100% sure but it should work. 
