A simple counterexample can be constructed using free objects:

For a set $X$ let $\mathbb{Z}\langle X\rangle$ be the free ring on $X$ and let $\mathbb{Z}F(X)$ be the group ring of the free group $F(X)$ on $X$. The inclusion $X \hookrightarrow F(X)$ induces a ring homomorphism $i: \mathbb{Z}\langle X\rangle \to \mathbb{Z}F(X)$ which is an epimorphism. If $x\neq y$ are in $X$ then $xy^{-1} \in \mathbb{Z}F(X)$ and there is obviously no "polynomial" $f \in \mathbb{Z}\langle X\rangle$ such that $fxy^{-1} \in \mathbb{Z}\langle X\rangle$.

To see that $i$ is an epimorphism, note that $\mathbb{Z}F(X)$ is generated (as a ring)
by $x,x^{-1}\;(x \in X)$ and a unitary ring homomorphism $\varphi: \mathbb{Z}F(X) \to T$ satisfies
$\varphi(x^{-1})=\varphi(x)^{-1}$. Hence $\varphi$ is determined by its values on $X$.

**Update:** The property in question holds true for commutative rings.

*Proof:* At first assume this has already been proved for zero-dimensional local commutative rings. Let $R \le S$ be an epimorphic extensions of comm. rings and choose a minimal prime $\mathfrak{p}$ of $R$ (exists by Zorn's lemma). $R_\mathfrak{p}$ is a local ring of dimension $\text{ht}(\mathfrak{p})=0$. Since $R\setminus \mathfrak{p}$ is a multiplicative subset of $S$ we can localize and obtain a comm. diagramm
$$\begin{array}{ccc}
R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline
{\scriptstyle\text{epi}}\downarrow & & \downarrow\scriptstyle\text{epi} \newline
R_\mathfrak{p} & \xrightarrow[i_\mathfrak{p}]{} & (R\setminus \mathfrak{p})^{-1}S
\end{array}$$
Hence $i_\mathfrak{p}$ is epi and it's easy to see that $i_\mathfrak{p}$ is mono as well.

Let $s \in S$. By our assumption there are $r_i/t_i \in R_\mathfrak{p},\;r_1/t_1 \neq 0$ such that $\frac{r_1}{t_1}\frac{s}{1}=\frac{r_2}{t_2}$, i.e. there is $t \in R\setminus \mathfrak{p}$ with $(tt_2r_1)s=tt_1r_2 \in R$. Moreover $tt_2r_1 \neq 0$ (because $r_1/t_1 \neq 0$ in $R_\mathfrak{p}$ just says there is no $t \in R\setminus \mathfrak{p}$ such that $tr_1=0$ in $R)$ and we are done.

Now suppose $R$ is zero-dimensional local comm. with max. ideal $\mathfrak{m}$. Again from the comm. diagramm
$$\begin{array}{ccc}
R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline
{\scriptstyle\text{epi}}\downarrow\;\; & & \downarrow\scriptstyle\text{epi} \newline
R/\mathfrak{m} & \xrightarrow[i_\mathfrak{m}]{} & S/\mathfrak{m}S
\end{array}$$
we conclude that $i_\mathfrak{m}$ is epi and since $R/\mathfrak{m}$ is a field, $i_\mathfrak{m}$ is an isomorphism (see the link in the OP's question). Hence each $s \in S$ can be written as
$$s=r + \sum_{i=1}^l m_is_i\qquad (r\in R, s_i \in S, m_i \in \mathfrak{m}, m_i \neq 0)$$
We want to show that there is $r_0 \in R,\;r_0 \neq 0$ such that $r_0s\in R$.
If $l=0$ then $s=r\in R$ and we are done. Otherwise, since $\mathfrak{m}=\sqrt{0}$ there is $n_1> 0$ maximal such that $m_1^{n_1} \neq 0$ and multiplying yields
$$m_1^{n_1}s=m_1^{n_1}r + \sum_{i=2}^l (m_1^{n_1}m_i)s_i.$$
If $m_1^{n_1}m_2=0$ ignore the corresponding summand. Otherwise, there is $n_2>0$ maximal such that $m_1^{n_1}m_2^{n_2}\neq 0$. Multiplying again yields
$$m_1^{n_1}m_2^{n_2}s=m_1^{n_1}m_2^{n_2}r + \sum_{i=3}^l(m_1^{n_1}m_2^{n_2}m_i)s_i.$$
Proceeding this way, we obtain the required $r_0$ in the form $r_0 = \prod_{j=1}^k m_{i_j}^{n_{i_j}}.\;\;$ qed.

Lectures on Modules and Rings, Example (1.4)), and then $id: R = S$ is a counter-example, right? – Torsten Schoeneberg Feb 11 '13 at 14:27nottrue for all epimorphic extensions: Consider $\mathbb{Z} \hookrightarrow \mathbb{Q} \times \mathbb{Z}/2$). (It is true e.g. for localisations.) – Torsten Schoeneberg Mar 4 '13 at 15:18