No for cardinality reasons.

Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf

Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$.

[note: I am not an expert and have not checked.]

[Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite-dimensional $R/I$-vector space, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]

No for cardinality reasons.

Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf

Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$.

[note: I am not an expert and have not checked.]

[Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite $R/I$-module, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]

No for cardinality reasons.

Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf

Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$.

[note: I am not an expert and have not checked.]

No for cardinality reasons.

Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf

Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $ \kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$.

[note: I am not an expert and have not checked.]

[Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite-dimensional $R/I$-vector space, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]