The answer to the question in the title is yes, as explained in the last paragraph below.

However, under a literal interpretation of "can" (implying actual feasibility), I believe the answer to the question in the body of the text is no.

Assume for instance that $f$ and $g$ are two is $p$-ordinary eigencuspforms ($p$-ordinary means that under a fixed embedding of $\bar{\mathbb Q}$ into $\bar{\mathbb Q}_{p}$, the $p$-adic valuation of $a_{p}(f)$ and of $a_{p}(g)$ is zero), that $\pi(f)_p$ (the automorphic representation of $\operatorname{GL}_{2}(\mathbb Q_{p})$ attached to $f$) is unramified principal series and that $\pi(g)_{p}$ (same notation) is unramified Steinberg.

Then the conductor of $f$ at $p$ is trivial ($r=0$) whereas the conductor at $p$ of $g$ is $p$ ($r=1$). However, after restriction to $I_{p}$, both $\rho_f$ and $\rho_g$ are equivalent to
\begin{equation}
\begin{pmatrix}
1&*\\
0&\chi^{-1}
\end{pmatrix}
\end{equation}
where $\chi$ is the cyclotomic character. I don't know how to distinguish between them using the class of the extension of $\chi^{-1}$ by $1$ (the $*$, so to speak) and it seems hard to me though I admit I also don't know that it is definitely not possible.

One can construct many such examples of ambiguous $I_{p}$-representation, so I doubt one can reconstruct $p^{r}$ in general. As more generally the representation $\rho_f|G_{\mathbb Q_{p}}$ is the representation $V_{2,a_p}$ in the notation of C.Breuil *Sur quelques représentations modulaires et $p$-adiques de $\operatorname{GL}_{2}(\mathbb Q_{p})$ II* (Journal de l'IMJ, 2003) it might be a good idea to have a look at this article if you want a definite answer.

As Aurel points out, $p^{r}$ is the conductor of the Weil-Deligne representation attached to $D_{\operatorname{pst}}(\rho_f|G_{\mathbb Q_{p}})$ so you certainly *can* reconstruct $r$ from $\rho_{f}|G_{\mathbb Q_{p}}$ and what you are missing in your setting are the eigenvalues of the image of $\operatorname{Fr}(p)$ through $\rho_f$. In the case above for instance, both eigenvalues would have the same $p$-adic valuations in the first case and different valuations in the second.