Can a nonzero polynomial evaluate to the zero function in a suitable infinite ring of char 0? I shall assume all rings to be commutative in this question. The impatient can scroll down to the "blockquote" to read the actual question.
Whenever we have a polynomial over a ring, it defines a function from the ring to itself by evaluation. It's reasonable to ask when two different polynomials define the same function.
From the factor theorem it follows that an $n^\text{th}$ degree polynomial over an integral domain has at most $n$ roots. Then it's easy to show this:
Theorem. Let $R$ be an infinite integral domain and let $f \in R[X]$ such that $f(a)=0$ for all $a \in R$, then $f = 0$.
Proof. $f$ has infinitely many roots, so it must be the zero polynomial. $\quad\square$
For finite rings a kind of opposite situation occurs:
Theorem. For any finite ring $R$ there are polynomials over $R$ that are different but agree on all elements.
Proof. There are only finitely many functions from $R$ to itself, but $R[X]$ is infinite. $\quad\square$
If we make further assumptions it's of course possible to prove more, as Pete L. Clark wrote in this post: [1]
Then there is the question of infinite rings that are not integral domains. It's relatively easy to come up with examples of a ring $R$ with positive characteristic and a nonzero polynomial that evaluates to the zero function, e.g.:
$$ R := \bigoplus_{n=1}^\infty \mathbb{Z}/6\mathbb{Z} \quad\text{and}\quad f(X) := X^3-X.$$
The Question:

This leaves open the case alluded to in this post's title: Is there a commutative ring of characteristic $0$ (hence infinite) such that a nonzero polynomial evaluates to the zero function?

 A: Consider the ring generated by $a$ with $a^2 = 0$, and take $p(x) = a x$.
A: Yes.  I'll give my example first.  Below is the TeXing I did while thinking that I was proving the answer to be "no".  Trying to prove the answer was "no" led me to this example:
Let $R=\mathbb{Z}[y]/\langle 6y,y^2\rangle$.  This commutative ring has characteristic zero, since no integer is in the ideal $\langle 6y,y^2\rangle$.  And now you can just slide over your polynomial example so that it always evaluates to zero:  $$f(X) = y\;X^3-y\;X=y\;(X^3-X)$$
Just as in your example, $X^3-X$ always evaluates to a multiple of $6$ when $X$ is an integer.  More generally if $X=a+b\;y$, then since $y^2$ is modded out, we only need consider the constant term $a$.
If you changed the question to be about integral domains rather than characteristic zero rings, then the answer would be "no" by completing the argument below.

Suppose that $f$ is such a polynomial in $R[x]$ of degree $n$: $$f(x)=\sum_{j=0}^n\;c_j\;x^j$$  The equations $$f(i)=0$$ for $i=0\ldots n$ form a system of $n+1$ linear equations in the unknowns $\{c_j\}$.  There is one clear solution to this system, where each $c_j=0$.  But can there be other solutions with $c_j\in R$?
The system can be written as 
$$\begin{bmatrix}1 & 0 & 0 & \cdots & 0\\
1 & 1 & 1 &\cdots & 1\\
1 & 2 & 4 &\cdots & 2^n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & n & n^2 & \cdots & n^n\end{bmatrix}
\begin{bmatrix}c_0\\c_1\\c_2\\\vdots\\ c_n\end{bmatrix}
=\begin{bmatrix}0\\0\\0\\\vdots\\0\end{bmatrix}$$
The matrix on the left (which I will call $V$) is an example of a Vandermonde matrix which is invertible in $M(\mathbb{Q})$.  Now, $V$ might not have an inverse in $M(R)$, but that's not a big problem.  It's still the case that in $M(R)$ there is a matrix $W$ such that $W\;V$ is a scalar matrix $D$ with an integer $d$ running down the diagonal.  You just need to rescale $V^{-1}$ by the least common multiple of the divisors that appear in $V^{-1}$. After applying $W$ to both sides, 
$$D
\begin{bmatrix}c_0\\c_1\\c_2\\\vdots\\ c_n\end{bmatrix}
=\begin{bmatrix}0\\0\\0\\\vdots\\0\end{bmatrix}$$
So there is some nonzero integer $d$, such that for each $j$, we have that $d\cdot c_j=0$.
Here I realized the answer is actually "yes".
