The answer to the first question is no. Here is a proof.
Let $A$ be a commutative ring and $\mathfrak{m}$ be a maximal ideal of $A$. Let $k=A/\mathfrak{m}$. Suppose we have an isomorphism of $A$-modules $A^I \cong A^{(I)}$, where $A^{(I)} := \bigoplus_{i \in I} A$. Tensoring with $k$ over $A$ yields an isomorphism of $k$-vector spaces
\begin{equation}
\frac{A^I}{\mathfrak{m} \cdot A^I} \cong k^{(I)}
\end{equation}
where $\mathfrak{m} \cdot A^I$ denotes the $A$-submodule generated by the $m\cdot x$ with $m \in \mathfrak{m}$ and $x \in A^I$. Since $\mathfrak{m} \cdot A^I \subset \mathfrak{m}^I$, we have a surjective $k$-linear map
\begin{equation}
\frac{A^I}{\mathfrak{m} \cdot A^I} \to k^I
\end{equation}
So we get a contradiction by considering the dimension over $k$ (and using the case you first settled).
EDIT : The answer to the second question is also negative in general. Let me prove that for $I=\mathbf{N}$, there exist a couple $(A,\mathfrak{m})$ such that the statement is false. The idea is to take $A=k[X_j, j \in J]$ and $\mathfrak{m}=\langle X_j, j \in J \rangle$ with a sufficiently big set $J$ (the argument should also work if $\mathfrak{m}/\mathfrak{m}^2$ is a sufficiently big $k$-vector space). We want to prove that
\begin{equation}
\dim \frac{A^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}.
\end{equation}
It is sufficient to show that $\dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}$. We have
\begin{equation}
\dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} \geq \dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}} + (\mathfrak{m}^2)^{\mathbf{N}}}
\end{equation}
The latter space can be identified with the cokernel of the natural (injective) map $k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$, where $V=\mathfrak{m}/\mathfrak{m}^2$.
Lemma The cokernel of $f : k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$ has dimension $\geq \dim V$.
Proof. A basis of $V$ is given by the family $e_j = \overline{X_j}$ indexed by $j \in J$. Write $J$ as a disjoint union $J = \sqcup_{s \in S} J_s$ where each $J_s$ is countable. We have $\operatorname{card} S = \operatorname{card J}$. Fix isomorphisms $\phi_s : \mathbf{N} \to J_s$. For any $s \in S$, let
\begin{equation}
u^{(s)} = (e_{\phi_s(n)})_{n \in \mathbf{N}} \in V^{\mathbf{N}}.
\end{equation}
Let us say that a sequence $u \in V^{\mathbf{N}}$ has finite rank if the vector space generated by the $u_n$ is finite dimensional. Then the image of $f$ is the subspace of finite rank sequences in $V^{\mathbf{N}}$. It is not difficult to see that a finite linear combination $\sum_s \lambda_s u^{(s)}$ with all $\lambda_s \neq 0$ cannot have finite rank, whence the lemma.
It remains to take $J$ sufficiently big, for example we can take $J=2^{k^{\mathbf{N}}}$.
The answer to both questions is yes.
As a preliminary, let's prove that for any infinite-dimensional vector space $V$, that
- Lemma: $card(V) = card(k) \cdot \dim V$
Proof: Since $card(k) \leq card(V)$ and $\dim V \leq card(V)$, the inequality
$$card(k) \cdot \dim V \leq card(V)^2 = card(V)$$
is obvious. On the other hand, any element of $V$ is uniquely of the form $\sum_{j \in J} a_j e_j$ for some finite subset $J$ of (an indexing set of) a basis $B$ and all $a_j$ nonzero. So an upper bound of $card(V)$ is $card(P_{fin}(B)) \sup_{j \in P_{fin}(B)} card(k)^j$. If $B$ is infinite, then $card(P_{fin}(B)) = card(B) = \dim(V)$, and for all finite $j$ we have $card(k^j) \leq card(k)$ if $k$ is infinite, and $card(k^j) \leq \aleph_0$ if $k$ is finite, and either way we have
$$card(V) \leq \dim V \cdot \max\{card(k), \aleph_0\} \leq \dim V \cdot card(k)$$
as desired. $\Box$
The rest is now easy. Suppose $I$ is an infinite set, and suppose without loss of generality that $V_i$ is nontrivial for all $i \in I$. Put $V = \prod_{i \in I} V_i$. We have
$$\dim V \geq \dim k^I = card(k)^I \geq card(k)$$
where the equality is due to Erdos and Kaplansky. Therefore
$$\dim(V) = \dim(V)^2 \geq \dim V \cdot card(k) = card(V) = \prod_i card(V_i)$$
by the lemma above.
Best Answer
I'm guessing that $R[\prod_i G_i]$ might be obtained as the inverse limit you wrote in the category of Hopf algebras over $R$. Here the forgetful functor
$$U: HopfAlg_R \to Coalg_R$$
preserves and reflects limits, so it suffices to check the claim in the category of (cocommutative) coalgebras. The guess then is that, by some application of the principle that every coalgebra is the filtered colimit of its finite-dimensional subcoalgebras, that the limit in $Coalg_R$ picks up only functions $\prod_i G_i \to R$ of finite support.
See these slides for some hints on calculating limits of Hopf algebras.