The answer is no; you can't have both equal to 1 in the equation you gave, but you can if you introduce $R_0$, as follows.
Friedmann-Lemaitre-Robertson-Walker metric is
$$
{ ds^2 = -c^2 dt^2 + a^2(t) \left( \frac{dr^2}{1 - K_0 r^2} +
r^2\left(d\theta^2 + \sin^2(\theta) d\phi^2\right) \right) }
$$
where $K_0$ is a constant. $K_0$ is equal to the Gaussian curvature of space at the time when $a=1$. (The metric is called FLRW since as I understand it, the history here is such that Lemaitre has as much right to be named in this list of names as the others.)
If $K_0 \ne 0$ then we can introduce
$$
k = K_0 /\, |K_0| \\
\bar{r} = \sqrt{|K_0|} \, r \\
R = a / \sqrt{|K_0|}
$$
and obtain
$$
{ ds^2 = -c^2 dt^2 + R^2(t) \left( \frac{d\bar{r}^2}{1 - k \bar{r}^2} +
\bar{r}^2 \left(d\theta^2 + \sin^2(\theta) d\phi^2 \right) \right) }
$$
where now $k$ has one of the values $-1,0,1$. Finally there is nothing to stop us writing $R(t) = R_0 a(t)$ where $a(t_0)=1$. Then we have
$$
{ ds^2 = -c^2 dt^2 + R_0^2 a^2(t) \left( \frac{d\bar{r}^2}{1 - k \bar{r}^2} +
\bar{r}^2 \left(d\theta^2 + \sin^2(\theta) d\phi^2 \right) \right) . }
$$
You can now drop the bar if you only want to use this form.
