The answer is yes - let $T$ be a Lawvere theory and $Mod_T$ its category of models in $Set$. Then with $C= Fun(Mod_T,Set)$ and $X= U$, the forgetful functor to $Set$, you recover $T$ as the "endomorphism Lawvere theory" of $X$.

A sketch of this goes as follows: $\hom_C(X^n,X^k)= \hom(U^n, U^k)$, but $U^n$ is corepresentable by the free $T$-algebra on $n$ generators $A_n$, so that by the Yoneda lemma, $\hom(U^n,U^k)\cong U(A_n)^k\cong \hom_{(Mod_T)^{op}}(A_n,A_k)\cong \hom_T(n,k)$

The answer is yes for Lawvere theories which admit a "zero", that is, Lawvere theories in which there is a natural map $x\to y$ (natural in $x,y\in T$). This is equivalent to a map from the terminal $T$-algebra $*$ to the initial $T$-algebra; and happens for instance when the initial $T$-algebra is terminal which is the case for $R$-modules.

This is a consequence of the following general observation:

For any small finite-product category $C$ in which there exists a transformation $x\to y$, natural in $x,y\in C$, there is a product-preserving embedding from $C$ into $Mod_S$ for some $S$.

Applying this to $C= T$, we get the desired result, so let me explain the proof of that.

Fix any such $C$, and consider the object $ Fun^\times(C,Set)$ given by $F:= \prod_{c\in C} \hom(c,-)$. The full subcategory $S\subset Fun^\times(C,Set)$ spanned by $F^{\times n}, n \in \mathbb N$ is a Lawvere theory with distinguished object $F$, and I claim $C$ embeds in a product-preserving way into $Mod_S$ - the embedding is given by $C\to Fun^\times(S,Set), c\mapsto (G\mapsto G(c))$.

(in terms of algebras, I am claiming that for a given Lawvere theory $T$, $S$ can be chosen as $\mathbf{Lawv}(\prod_n Fn)$, where $F$ is the free $T$-algebra functor)

This is clearly a functor, so now it suffices to see that it is in fact fully faithful. A map $f: d\to d'$ induces $\prod_{c\in C}\hom(c,d)\to\prod_{c\in C}\hom(c,d')$. Since all the terms in the product are nonempty, it follows that one can recover the component $\hom(d,d)\to \hom(d,d')$ from this map and thus it is faithful.

Now, let $\alpha_G: G(d)\to G(d')$ be a morphism, natural in $G\in S$. I claim that it must come from some map $f :d\to d'$.

Fix some natural morphism $0_{x,y}: x\to y$. For any $d\in C$, consider the following morphism $F\to F$- $\prod_{c\in C}\hom(c,-)\to \prod_{c\in C}\hom(c,-)$ : on the $c$-coordinate of the target, pick out $0_{c,-}\in \hom(c,-)$ if $c\neq d$, and on the $d$-coordinate, just pick the projection $F\to \hom(d,-)$. This is a morphism of functors.

Naturality of $\alpha$ forces it to commute with this, and it follows that $\alpha_F$ is a product of maps $\hom(c,d)\to \hom(c,d')$. The claim is now that these are natural in $c$, and this can be proved in a similar way, by considering, for every $g:c_0\to c_1$ the map $F\to F$ which is the identity on most coordinates, and which on the coordinate $\hom(c_0,-)$ first projects onto $\hom(c_1,-)$ and then precomposes by $g$.

As it is natural in $c$, it forces it to be given by $f\circ -$ for some (unique) $f: d\to d'$.

Now for a general $G= F^{\times n}\in S$, you can use the projections $G\to F$ and naturality of $\alpha$ to deduce that the whole of $\alpha$ is in fact given by $G(f): G(d)\to G(d')$, which proves that our functor was, in fact, fully faithful.