I know that by conversion theorem, $ r → q = \lnot r \vee q $. So if $¬r → q$ will be equal to $r \vee q$ and not $r \vee ¬q$ right?
Let's say we know that $q$ and $p$ are true. Given that $ r → s; ¬r → q; p \vee r;$ are true.
How do we know that whether $r$ is true or false?
Given
$$\begin{array}\\ \lnot\ r\implies q\\ p\ \lor\ r\\ \hline (r\ \lor\ q)\ \land (p\ \lor\ r) \end{array}$$
Since you said $p$ and $q$ are true, whether or not $r$ is true is not important/ it won't affect the truth of the conclusion/ it's unknown/ it can be either.
And in your title it's $\large\lnot\ r\implies q \color{blue}{\equiv} r\ \lor\ q$, not $\color{red}{=}$.
If you want to prove that $r$ is false, try to find $t$ s.t.
$$\begin{array}\\ r\implies t\\ t\ \land\ s\equiv\bot\ ,\\ \end{array}$$
which $\bot$ means a contradiction, then
$$\begin{array}\\ r\implies s\\ r\implies t\\ t\ \land\ s\equiv\bot\\ \hline \lnot\ r \end{array}$$