$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$The desired answer clearly is meant for a practice in classical mechanics without relativistic effects.
Presumably the OP wants to see his/her method comes to fruition so this answer continues the approach in the question post. There exists analysis that some might deem more physical and nice (see this answer or that).
First of all, your original expression is off by a negative sign. It is true that $r$ decreases along time $t = 0$ to $t = t_{ff}$ so the derivative is negative (taking the minus square root). Accordingly, the integration limit should start at $r = R$ at $t = 0$ and ends at $r = 0$ at $t = t_{ff}$. Hence the negative sign cancels and we have an integration of positive square root from $r = 0$ to $R$.
$$\int \dd{t} = \frac1{ \sqrt{2GM} }\int^R_0 \left( \frac1r - \frac1R \right)^{\frac{-1}2} \dd{r} = \sqrt{\frac{R}{2GM} }\int^R_0 \left( \frac{R}r - 1 \right)^{\frac{-1}2} \dd{r} $$
Upon seeing "something minus one" in the square root, one should think of $\sec^2\theta$ for the very useful $\sec^2\theta - 1 = \tan^2\theta$.
To have $\frac{R}r = \sec^2\theta$ one considers the change of variable $\theta \equiv \arccos \bigl(\sqrt{ \frac{r}R } \bigr)$, a well-defined one-to-one mapping where the lower integration limit $r = 0$ becomes $\theta = \frac{\pi}2$ and at upper $r = R$ it is $\theta = 0$.
\begin{align}
\frac{R}r = \sec^2\theta & \implies \frac{-R}{r^2} \dd{r} = 2 \sec\theta \cdot \sec\theta\tan\theta \dd{\theta} \\
&\implies \frac{-1}R \bigl( \frac{R}r \bigr) ^2 \dd{r} = 2 \sec^2\theta\tan\theta \dd{\theta} \\
&\implies \frac{-1}R \sec^4\theta \dd{r} = 2 \sec^2\theta\tan\theta \dd{\theta} \\
&\implies \dd{r} = -2R \cos^2\theta\tan\theta \dd{\theta}
\end{align}
Meanwhile, the integrand $\sqrt{ \frac{R}r - 1} = \tan\theta$ is the positive root by definition. Putting things together:
\begin{align}
t_{ff} = \int_{t = 0}^{t_{ff} } \dd{t} &= \sqrt{\frac{R}{2GM} }\int^R_0 \frac1{ \sqrt{ \frac{R}r - 1} } \dd{r} \\
&= \sqrt{\frac{R}{2GM} }\int_{\theta = \frac{\pi}2 }^0 \frac1{\tan\theta} (-2R)\cos^2 \theta \tan\theta \dd{\theta} \\
&= R \sqrt{\frac{R}{2GM} }\int_{\theta = 0}^{ \frac{\pi}2 } 2\cos^2\theta \dd{\theta} \\
&= R \sqrt{\frac{R}{2GM} }\int_{\theta = 0}^{ \frac{\pi}2 } \bigl( \cos(2\theta) + 1 \bigr) \dd{\theta} \\
&= R \sqrt{\frac{R}{2GM} } \frac{ \sin(2\theta) + 2\theta }2\Bigg|_{\theta = 0}^{ \frac{\pi}2 } \\
& = R \sqrt{\frac{R}{2GM} } \frac{ \pi }2
\end{align}
Squaring this elapsed-time $t_{ff}$ yields the desired expression of $t_{ff}^2$ seen in the question post.$\endgroup$