Watson, the answer to your question is that those groups are nonisomorphic as topological groups, and by being a bit more careful they are nonisomorphic as abstract groups also. We will pass to the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, which is ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$, and show this group knows what $p$ is. First we will check that the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is a purely algebraic construct.
For a topological group, its abelianization is the quotient by its commutator subgroup, just like for abstract groups, but there is a catch: the definition of the commutator subgroup of a topological group is the closure of the ordinary commutator subgroup (viewing the group as an abstract group). Fortunately, it turns out that in the case of $\mathbf Q_p$, or finite extensions of $\mathbf Q_p$, the ordinary commutator subgroup of the absolute Galois group is closed. This is because of the following two facts.
Jannsen and Wingberg proved that the absolute Galois group of a finite extension $K/\mathbf Q_p$, for $p \not= 2$, is topologically finitely generated (in fact finitely presented): this is in a few articles in Inventiones vol. 70 (1982/83). Diekert settled the case $p=2$ shortly afterwards. Diekert gives an explicit finite presentation when $\sqrt{-1} \in K$ for a $2$-adic field $K$, but if $\sqrt{-1} \not\in K$ then by applying Diekert's result to $K(\sqrt{-1})$, an open subgroup of index $2$ in ${\rm Gal}(\overline{K}/K)$ is topologically finitely generated (in fact topologically finitely presented) so ${\rm Gal}(\overline{K}/K)$ is topologically finitely generated with one more generator.
Novikov and Segal (see here) showed that for every topologically finitely generated profinite group, its commutator subgroup as an abstract group is closed. (Serre had proved this earlier when the group is a topologically finitely generated pro-$p$ group.)
Putting these together, the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ as an abstract group is the same as its abelianization as a topological group, which is ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$. Local class field theory tells us that the group ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$ is isomorphic to $\widehat{\mathbf Z} \times \mathbf Z_p^\times$ as topological groups, and hence as abstract groups. We will show the abstract group $\widehat{\mathbf Z} \times \mathbf Z_p^\times$ knows what $p$ is.
Set $G = \widehat{\mathbf Z} \times \mathbf Z_p^\times$, so the torsion subgroup $G_{\rm tor}$ is $(\mathbf Z_p^\times)_{\rm tor}$, which is a finite subgroup of $\mathbf Z_p^\times$: $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathbf Z_p^\times)/(\mathbf Z_p^\times)_{\rm tor}$. The group $\mathbf Z_p^\times$ is a topological group, whether you want to think of it like that or not. Using the $p$-adic logarithm (extended from $1 + p\mathbf Z_p$ to $\mathbf Z_p^\times$ by setting it to be $0$ on Teichmuller roots of unity), we have $\mathbf Z_p^\times/(\mathbf Z_p^\times)_{\rm tor} \cong \log(1+p\mathbf Z_p)$ as topological groups, and the image of the $p$-adic logarithm on $1+p\mathbf Z_p$ is $p\mathbf Z_p$ for $p > 2$ and $4\mathbf Z_2$ for $p = 2$. In both cases the image is isomorphic to $\mathbf Z_p$, so $\mathbf Z_p^\times/(\mathbf Z_p^\times)_{\rm tor} \cong \mathbf Z_p$ as topological groups and thus also as abstract groups. Therefore
$$G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathbf Z_p^\times)/(\mathbf Z_p^\times)_{\rm tor} \cong \widehat{\mathbf Z} \times \mathbf Z_p \cong \left(\prod_{\ell} \mathbf Z_\ell\right) \times \mathbf Z_p \cong
\prod_{\ell \not= p} \mathbf Z_\ell \times \mathbf Z_p^2$$
as abstract groups. To show the abstract group $\widetilde{G} := G/G_{\rm tor}$ knows $p$, observe that for a prime number $\ell$, $\widetilde{G}/\ell\widetilde{G}$ is isomorphic to $\mathbf Z/\ell\mathbf Z$ if $\ell \not= p$ and is isomorphic to $(\mathbf Z/p\mathbf Z)^2$ if $\ell = p$, so $|\widetilde{G}/\ell\widetilde{G}| \not= \ell$ only at $\ell = p$. Thus $G$, as an abstract group, knows what $p$ is.
Next you will probably ask, for $K$ a finite extension of $\mathbf Q_p$ and $L$ a finite extension of $\mathbf Q_q$, where $p$ and $q$ are different primes, could ${\rm Gal}(\overline{K}/K)$ be isomorphic to ${\rm Gal}(\overline{L}/L)$ as abstract groups? If they were, then again we could pass to the abelianizations (abstract and topological abelianizations are the same thing by the two hard theorems at the start) and local class field theory then implies $\widehat{\mathbf Z} \times \mathcal O_K^\times \cong \widehat{\mathbf Z} \times \mathcal O_L^\times$ as abstract groups. We want to show the abstract group $G := \widehat{\mathbf Z} \times \mathcal O_K^\times$ knows what $p$ is, where $K$ is a finite extension of $\mathbf Q_p$.
The solution we used when $K = \mathbf Q_p$ can be made to work when $K$ is a finite extension of $\mathbf Q_p$, with some further details needed. The torsion subgroup $G_{\rm tor}$ of $G$ equals $(\mathcal O_K^\times)_{\rm tor}$ and $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathcal O_K^\times)/(\mathcal O_K^\times)_{\rm tor}$ as abstract groups. Using the $p$-adic logarithm (extended from $1 + \mathfrak m_K$ to $\mathcal O_K^\times$ by killing the Teichmuller roots of unity in $\mathcal O_K^\times$), $(\mathcal O_K^\times)/(\mathcal O_K^\times)_{\rm tor}$ is isomorphic to $\log(1 + \mathfrak m_K)$. We'd like to say $\log(1 + \mathfrak m_K)$ is isomorphic to $\mathcal O_K$ as topological groups, but this requires some extra care compared to the case $K = \mathbf Q_p$ because $\log(1 + \mathfrak m_K)$ need not be inside $\mathcal O_K$! See the MO page here for some explicit examples of that in (highly ramified) $p$-power cyclotomic extensions of $\mathbf Q_p$.
Claim: For every finite extension $K$ of $\mathbf Q_p$, $\log(1 + \mathfrak m_K) \cong \mathcal O_K$ as topological groups.
Proof of claim: Since $1+\mathfrak m_K$ is a $\mathbf Z_p$-module by exponentiation, $\log(1+\mathfrak m_K)$ is a $\mathbf Z_p$-module in $K$ (using ordinary multiplication by $\mathbf Z_p$). We will show $\log(1+\mathfrak m_K)$ both contains and is contained in $\mathbf Z_p$-modules isomorphic to $\mathcal O_K$. Since $\mathcal O_K$ is a finite free $\mathbf Z_p$-module of rank $n$, where $n = [K:\mathbf Q_p]$, we get $\log(1+\mathfrak m_K) \cong \mathbf Z_p^n$ as $\mathbf Z_p$-modules from the structure theory for finitely generated modules over a PID (such as $\mathbf Z_p$). That implies $\log(1 + \mathfrak m_K) \cong \mathcal O_K$ as $\mathbf Z_p$-modules and thus as topological groups.
To show $\log(1+\mathfrak m_K)$ contains a $\mathbf Z_p$-module isomorphic to $\mathcal O_K$, let $\pi$ be a uniformizer of $K$. On a small enough disc around $1$ (of radius less than $(1/p)^{1/(p-1)}$) the $p$-adic logarithm is an isometry with inverse being the $p$-adic exponential, so $\log(1 + \pi^N\mathcal O_K) = \pi^N\mathcal O_K$ for $N \gg 0$. (The values of the $p$-adic logarithm are inside $K$ since the logarithm is a power series with rational coefficients.) Therefore $\log(1+\mathfrak m_K)$ contains a $\mathbf Z_p$-module isomorphic to $\mathcal O_K$.
To show $\log(1+\mathfrak m_K)$ is contained in a $\mathbf Z_p$-module isomorphic to $\mathcal O_K$, by compactness of $1+\mathfrak m_K$ the image $\log(1+\mathfrak m_K)$ is compact in $K$, so it is contained in $(1/\pi^M)\mathcal O_K$ for $M \gg 0$. Therefore $\log(1+\mathfrak m_K)$ is contained in a $\mathbf Z_p$-module isomorphic to $\mathcal O_K$. This completes the proof of the claim.
Note: It was crucial that we looked at the $p$-adic logarithm on $1+\mathfrak m_K$ for $K$ a finite extension of $\mathbf Q_p$: if $K = \mathbf C_p$ then $\log(1+\mathfrak m_K) = \mathbf C_p$! We don't need that so I don't discuss this further.
Using the claim, $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathcal O_K^\times)/(\mathcal O_K^\times)_{\rm tor} \cong \widehat{\mathbf Z} \times \mathbf Z_p^n$ as topological groups, where $n = [K:\mathbf Q_p]$, and thus also we have an isomorphism as abstract groups. (I am using topology merely as a tool to understand the structure of $G/G_{\rm tor}$ as an abstract group.)
Setting $\widetilde{G} := G/G_{\rm tor}$, for each prime number $\ell$ we have $\widetilde{G}/\ell\widetilde{G} \cong \mathbf Z/\ell\mathbf Z$ if $\ell \not= p$ and $\widetilde{G}/p\widetilde{G} \cong (\mathbf Z/p\mathbf Z)^{n+1}$, with $n+1 \geq 2$, so as before, $|\widetilde{G}/\ell\widetilde{G}| \not= \ell$ only at $\ell = p$.
Hence the abstract group $\widehat{\mathbf Z} \times \mathcal O_K^\times$, where $K$ is a finite extension of $\mathbf Q_p$, knows what $p$ is.
Best Answer
There's indeed no continuous surjective homomorphism $\hat{\mathbf{Z}}^\times\to\mathbf{Z}_p^2$ for any prime $p$.
Indeed, we have $\hat{\mathbf{Z}}^\times\simeq\prod_p(\mathbf{Z}_p\times\mathbf{Z}/q_p\mathbf{Z})$. Any homomorphism into $\mathbf{Z}_p$ has to be trivial on $\mathbf{Z}_\ell$ for any $\ell\neq p$ and on $\mathbf{Z}/q_p\mathbf{Z}$ for all $p$. By continuity, it therefore factors through $\mathbf{Z}_p$. Since there's no surjective continuous homomorphism $\mathbf{Z}_p\to\mathbf{Z}_p^2$, we're done.