The following partial answer (inspired by Pace Nielsen's deleted answer) addresses only the instances where $|R|=\aleph_0$ and $|S|\gt2^{\aleph_0}$. I claim that it's consistent (relative to the consistency of a measurable cardinal) that Alice has a winning strategy in these cases.
Given an infinite cardinal $\kappa$, let $G(\kappa)$ denote the game played by Alice and Bob on $R\times S$ where $R=\mathbb N$ and $|S|=\kappa$, but with the payoff modified in Bob's favor as follows: The set $S$ is partitioned into disjoint sets $S_1$ and $S_2$ with $|S_1|=|S_2|=\kappa$; Alice wins iff there exist $s_1\in S_1$ and $s_2\in S_2$ such that, for each $r\in R$, at least one of $(r,s_1)$ and $(r,s_2)$ is colored red. Plainly, if Alice wins this "bipartite" game, then she also wins the original game.
In fact I will show that Alice has a winning strategy in the game $G(\kappa)$ on the assumption that White has a winning strategy in a certain game $U(\kappa)$ which was suggested (in a slightly different form) on pp. 346–347 of S. Ulam, Combinatorial analysis in infinite sets and some physical theories, SIAM Review 6 (1964) 343–355, and discussed by F. Galvin and M. Scheepers, Baire spaces and infinite games, Arch. Math. Logic 55 (2016), 85–104.
The game I call $U(\kappa)$ is $\mathbf{UG}(\aleph_0,\kappa)$ in the notation of G & S, whose exposition of unpublished work of J. Baumgartner, C. Gray, R. Laver, M. Magidor, and R. Solovay (the history is summarized in Section 4 of the paper) includes the following results:
Theorem 11. White has no winning strategy in $U(2^{\aleph_0})$.
Theorem 17. If it's consistent that there is an uncountable measurable cardinal, then it's consistent that $2^{\aleph_0}=\aleph_1$ and White has a winning strategy in $U(\aleph_2)$.
Definition. The game $U(\kappa)$ is played on a set $E$ of cardinality $\kappa$. First White partitions $E$ into $\aleph_0$ pieces (some of which may be empty) and Black chooses one of those pieces, call it $B_1$. Then Black partitions $B_1$ into $\aleph_0$ pieces and White chooses one of them, call it $W_1$. Then White partitions $W_1$ into $\aleph_0$ pieces, and so on. White wins if the sequence $B_1\supseteq W_1\supseteq B_2\supseteq W_2\supseteq\cdots$ (of length $\omega$) has a nonempty intersection.
Theorem. If White has a winning strategy in $U(\kappa)$, then Alice has a winning strategy in $G(\kappa)$.
Proof. We may assume that the game $G(\kappa)$ is played in the following way: in the $n^\text{th}$ inning, first Alice chooses a function $f_n:S\to\mathbb N$ and colors red all uncolored points $(r,s)$ with $r\le\max\{n,f_n(s)\}$, and then Bob chooses a function $g_n:S\to\mathbb N$ and colors blue all uncolored points $(r,s)$ with $r\le g_n(s)$. (So everything will be colored after $\omega$ moves.)
Alice's strategy works independently on $S_1$ and $S_2$. To keep the notation simple I'll describe what happens on $S_1$; the play on $S_2$ is similar. Let $\sigma$ be a winning strategy for White in $U(S_1)$, the game $U(\kappa)$ played on the set $S_1$. As the game $G(\kappa)$ proceeds, Alice will imagine a branching tree of $\sigma$-plays of $U(S_1)$.
First, Alice partitions $S_1$ into $\aleph_0$ pieces $P_1,P_2,\dots$ according to the strategy $\sigma$, and then chooses an unbounded function $f_1:S_1\to\mathbb N$ which is constant on each piece. (Of course $P_i\ne\varnothing$ since $\sigma$ is a winning strategy.) We don't know which piece Black will choose, so each $P_i$ becomes a "parallel world"; further play on $P_i$ proceeds as if $P_i$ had been chosen by Black in $U(S_1)$.
Next, Bob chooses a function $g_1:S_1\to\mathbb N$. Alice imagines that Black has partitioned each P_i into $\aleph_0$ smaller pieces $P_{i,j}$ on which $g_1$ in constant. For each $i$ she chooses a piece $P_{i,j}$ according to the strategy $\sigma$. From now on Alice doesn't care what happens on the unchosen pieces, but she partitions each chosen piece $P_{i,j}$ into $\aleph_0$ pieces $P_{i,j,k}$ according to $\sigma$, and so forth.
I've described a strategy for Alice in $G(\kappa)$. To show that Alice wins, I will construct for each $i\in\{1,2\}$ a $\sigma$-play of $U(S_i)$ and let $s_i$ be a point in the (nonempty) intersection of the chosen sets.
First, let's make Black's first move in $G(S_1)$ by choosing (arbitrarily) one of the pieces $P_i$ from White's initial partition; call the chosen piece $B^1_1$. After Black chose $B^1_1$ he partitioned it into $\aleph_0$ pieces (level sets of $g_1|B^1_1$) and White chose one of them which we call $W^1_1$. Now we have $B^1_1\supseteq W^1_1$; $f_1$ is constant on $B^1_1$ and $g_1$ is constant on $W^1_1$.
Next we make Black's first move in $G(S_2)$, choosing $B^2_1$ so that $f_1(B^2_1)\ge g_1(W^1_1)$, thus ensuring for all $s_1\in W^1_1$ and $s_2\in B^2_1$, and for all $r\in\mathbb N$, that if $(r,s_1)$ was colored blue on Bob's first turn, then $(r,s_2)$ was colored red on Alice's first turn.
Next Black partitioned $B^2_1$ into $\aleph_0$ pieces and White chose one them, call it $W^2_1$, on which $g_1$ is constant. Going back to $S_1$, White partitioned $W^1_1$ into $\aleph_0$ pieces on which $f_2$ is constant; we make Black's next move in $U(S_1)$, choosing a piece $B^1_2\subseteq W^1_1$ with $f_2(B^1_2)\ge g_1(W^2_1)$, thus ensuring for all $s_1\in B^1_2$ and $s_2\in W^2_1$ and for all $r\in\mathbb N$, that if $(r,s_2)$ was colored blue on Bob's first turn, then it was colored red on Alice's second turn, if not sooner.
Continuing in this way, we construct plays of $G(S_1)$ and $G(S_2)$ which are won by White, so we can find a point $s_1\in B^1_1\cap W^1_1\cap B^1_2\cap\cdots$ and a point $s_2\in B^2_1\cap W^2_1\cap B^2_2\cap\cdots$, and for each $r\in\mathbb N$ at least one of the points $(r,s_1)$ and $(r,s_2)$ will be colored red.
