For arbitrary RR(1) grammar $G$, is there an equivalent LL(1) grammar $G'$?
No. A language $\mathcal{L}$ is $RR(k)$ exactly if its reverse $\mathcal{L}^R$ is $LL(k)$. So if a $RR(k)$ grammar $G$ could be transformed into an $LL(k)$ grammar $G'$, that would mean that both $L(G^R)$ and $L(G') = L(G)$ are $LL(k)$. 

But $LL(k)$ is not closed under reversal; there are $LL(k)$ languages whose reverse is no $LL(k)$. The classic demonstration that the set of $LL(1)$ languages is a proper subset of the set of $LR(1)$ languages also serves for this question.

The language $\{a^ib^j\mid 0 \le i\le j\}$ is $LL(1)$:

$$\begin{align}
S &amp;\to A B\\
A &amp;\to \epsilon \\
A &amp;\to a A b\\
B &amp;\to \epsilon \\
B &amp;\to b B
\end{align}$$

However, its reverse $\{b^ja^i\mid 0 \le i \le j\}$ is not $LL(k)$ for any $k$, although the reverse is clearly $RR(1)$, since a n $RR(1)$ grammar for the reverse language can be constructed by simply reversing every production in the original grammar:

$$\begin{align}
S &amp;\to B A\\
A &amp;\to \epsilon \\
A &amp;\to b A a\\
B &amp;\to \epsilon \\
B &amp;\to B b
\end{align}$$

(As it happens, the above grammar is LR(1). The language is an abstraction of the "dangling else" syntax.)