$$R_{s}=\mathbb{E}[R_{t}|S_{t}=s]$$
is the expected reward at time step $t$ given that the state at time $t$ is $s$, where
- $R_{t}$ and $S_t$ are random variables that represent the reward and state at time $t$, respectively,
- $S_{t}=s$ is an event, and
- $\mathbb{E}[X]$ denotes the expected value of the random variable $X$.
It does not matter how you entered $S_t = s$. The only thing that matters is that the current state is $s$. So, the answer to your question in the title is - no. $R_s$ is defined as an expectation (average) of the reward for being in a state, and, in this case, it doesn't take into account next states.
Suppose that the reward space $\mathcal{R} \subset \mathbb{R}$ is a discrete set, then we can write $R_s$ as follows
$$R_{s}=\sum_{r \in \mathcal{R}} r p(r \mid S_{t}=s),$$
where $p(r \mid S_{t}=s)$ is a conditional probability distribution that describes how the reward is distributed in a given state. If you always get the same reward, i.e. $p(r \mid S_{t}=s) = 1$ for a particular $r$ and $0$ for all others, then you have a deterministic reward function, i.e. $R_s$ is equal to the $r$ for which $p(r \mid S_{t}=s) = 1$. In most RL examples, reward functions are deterministic.
Note that it is probably more common to define the reward function as a function of the state $S_t = s$ and action taken in $s$, $A_t =a$, and write it as $R(s, a)$ or $r(s, a)$, i.e.
\begin{align}
R(s, a)
&=\mathbb{E}[R_{t}|S_{t}=s, A_t = a]\\
&=\sum_{r \in \mathcal{R}} r p(r \mid S_{t}=s, A_t = a)
\end{align}
Note that we're now using $p(r \mid S_{t}=s, A_t = a)$ and not $p(r \mid S_{t}=s)$.
You could also define $R(s, a, s')$. See this or Sutton & Barto's book for more details.
