Under a literal interpretation of "can" (implying actual feasibility), I believe the answer 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.

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.

Under a literal interpretation of "can" (implying actual feasibility), I believe the answer 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.

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 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.

Under a literal interpretation of "can" (implying actual feasibility), I believe the answer 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.