Digitally undoing a hardware filter
Using the Euler approximation, the mapping between the $s$-plane and the $z$-plane would be:

$$
  s \rightarrow \frac{z - 1}{T}
$$
where $1/T = f_s = 30$ (thus $s \rightarrow 30(z-1)$).

Substituting back in the original transfer function yields:

$$
\begin{align}
  H(z) &amp;= \frac{24.2\cdot 30 (z-1) + 1}{2.2\cdot 30 (z-1) + 1} \\
       &amp;= \frac{726 (z-1) + 1}{66 (z-1) + 1} \\
       &amp;= \frac{726z - 725}{66z-65} \\
       &amp;= \frac{726 - 725z^{-1}}{66-65z^{-1}}
\end{align}
$$

Which converted to a difference equation is:

$$
V_{in}[n] \approx 0.0909V_{out}[n] - 0.0895V_{out}[n-1] + 0.9986V_{in}[n-1]
$$

As far as whether retrieving the original signal is possible partly depends on whether the original signal did not in fact include significant energy above $f_s/2$.