Module category equivalent to graded module category? Main Question
Let $R$ be a graded ring, graded by the nonnegative integers.  Denote by $\mathrm{gr}R-\mathrm{Mod}$ the category of $\mathbb{Z}$-graded left $R$-modules with morphisms that preserve the grading.  Is there a ring $S$ such that the category $S-\mathrm{Mod}$ of left $S$-modules is equivalent to $\mathrm{gr}R-\mathrm{Mod}$?
As the category of graded $R$-modules is abelian, the Freyd-Mitchell theorem guarantees an exact embedding into a module category, but this is not necessarily an equivalence of categories, right?

Motivation
My motivation for the question is an offhand remark made to me indicating that for a given ring $A$, there is a ring $S$ such that the category of complexes of $A$-modules is equivalent to the category of $S$-modules.  If you define the graded ring $R = A[t]/(t^2)$, graded by powers of $t$, then (I think) complexes of $A$-modules are equivalent to graded $R$-modules, so the question is reduced to the main question stated above.
Of course, it could be the case that the answer to the original question I asked is negative, and yet the offhand remark is still true, in which case I would be interested in hearing about why that is.

Edit
I neglected to mention that I was hoping for a unital ring $S$ with unital modules.  As several people have pointed out in the comments, this is not possible.  I thought I would put up an argument to show why this is in case people come looking at this post in the future.
Theorem 1 of Chapter 4, Section 11 of the book Categories and Functors, by Bodo Pareigis, gives a complete characterization of when an abelian category $\mathcal{C}$ is a module category.  The criteria are that $\mathcal{C}$ must contain a progenerator (i.e. a finite, projective generator; I had to look that up) and it must contain arbitrary coproducts of that generator.
Now let's see that $\mathrm{gr}R-\mathrm{Mod}$ cannot contain a finite progenerator.  Take any finite (hence finitely generated) projective module $P = \bigoplus_{n \in \mathbb{Z}} P_n$.  Since $P$ is finitely generated, there is some index $k_0$ such that $P_n = 0$ for $n<k_0$ (this uses the fact that $R$ is graded by the nonnegative integers).
It is the fact that all components of $P$ below $k_0$ vanish that prevents $P$ from being a generator.  For example, take the graded module $M$ such that $M_{k_0 -1} = R_0$ and all other $M_n = 0$.  Since the only map from $P$ to $M$ is the zero map, morphisms from $P$ to $M$ cannot distinguish morphisms originating from $M$.  Hence $P$ cannot be a generator.
I accepted Mariano's answer because I felt it was the most elegant, but I learned something from all of the answers posted.  Thanks everyone!
 A: $\newcommand\ZZ{\mathbb{Z}}$Let $R=\bigoplus_{n\in\mathbb N_0}R_n$ be your graded ring. Construct a category $Q$ with objects $\ZZ$ and where $\hom_Q(n,m)=R_{m-n}$ with composition coming from the multiplication in $R$. A graded left $R$-module is the same thing as a functor from $Q$ to abelian groups. The (non-unital) ring associated to $Q$ does what you want.
A: This answer is just an elaboration of Moosbrugger's comment.
For simplicity, assume that $R$ is just a field $k$ concentrated in degree zero.  Put $S=\text{Map}(\mathbb{Z},k)$, and let $e_n\in S$ be the obvious idempotent supported at $n$.  For any $S$-module $M$ we can put $(FM)_n=M/((1-e_n)M)$; this gives a functor $F:\text{Mod}_S\to\text{GrMod}_R$.  Some comments above suggest that this should be an equivalence, but it is not.  To see this, let $J\leq S$ be the ideal of finitely-supported functions.  It is then easy to see that $S/J\neq0$ but $F(S/J)=0$.  This is the point behind Moosbrugger's remark about topologies.  Suppose we give $S$ the product topology, and consider only modules $M$ such that the action map $S\times M\to M$ is continuous with respect to the discrete topology.  I think this just means that for all $m\in M$ there is a finite set $K$ such that $m=\sum_{k\in K}e_km$.  Using this we find that $M\simeq\bigoplus_{n\in\mathbb{Z}}(FM)_n$, and thus that $F$ gives an equivalence from the continuous module category to $\text{GrMod}_R$.
A: The answer is yes. Let $T'$ be the path $\mathbb{Z}$-algebra of the Quiver
$$\cdots\rightarrow n-1\rightarrow n \rightarrow n+1 \rightarrow\cdots$$
This $\mathbb{Z}$-algebra is defined by generators $\{a_n,b_n\}_{n\in\mathbb{Z}}$, where $a_n$ is morally the $n^{\text{th}}$ vertex of the quiver, and $b_n$ is the morphism $n\rightarrow n+1$. The derining relations are
$$a_n^2=a_n,\qquad b_na_n=b_n, \qquad a_{n+1}b_n=a_n,\qquad a_ma_n=0\;\text{ if }\;m\neq n.$$
A right $R\otimes T'$-module $M$ is the same as a diagram of $R$-modules
$$\cdots\rightarrow M_{n-1}\rightarrow M_n \rightarrow M_{n+1} \rightarrow\cdots$$
The correspondence is given by $M_n=a_nM$, and $M_n \rightarrow M_{n+1}$ is left multiplication by $b_n$.
Since you want complexes, let $T$ be the quotient of $T'$ by the additional relations
$$b_{n+1}b_n=0.$$
Right $R\otimes T$-modules are complexes of $R$-modules.
You may dislike that $T'$, $R\otimes T'$, $T$ and $R\otimes T$ don't have a unit. But this is easy to solve. Just take the unitalization of these rings ($\mathbb{Z}$-algebras) and unital modules over them.
