Integrating radial free fall in Newtonian gravity

If $h$ is the height about the earth then

$$ \ddot{h} = -\frac{G M}{(R+h)^2} $$

$$ \ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h} $$

$$ \int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$

$$ \int -\frac{G M}{(R+h)^2}\; {\rm d} h = \frac{1}{2} \dot{h}^2 + K_1 $$

$$ \frac{G M}{R+h} = \frac{1}{2} \dot{h}^2 + K_1 $$

Given initial velocity of 0 at a height $h_0$ then $K_1=\frac{G M}{R+h_0}$ and

$$ \dot{h} = \sqrt{ \frac{2 G M (h_0-h)}{(R+h)(R+h_0)}} $$ gives the velocity profile as a function of height $h$. The time to distance is

$$ t = \int \frac{1}{\dot{h}}\;{\rm d}h + K_2 $$

which can be expressed as

$$t \sqrt{ \frac{2 G M}{(R+h_0)^3} } = \cos^{-1}\left( \sqrt{ \frac{R+h}{R+h_0}}\right) - \sqrt{ \frac{r+h}{R+h_0} \left( 1 - \frac{R+h}{R+h_0} \right) } $$

A close approximation of the above is

$$ h \approx (R+h_0)\left(1-\left( \frac{9 G M}{2 r_0^3} t^2 \right)^\frac{1}{3} \right) - R $$

Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$ E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)} $$

For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order differential which can be solved by separating the variables.

So you were on the right track with integrating over r and over t. Here's how you could do it:
The acceleration at any radius, r (if we assume Earth is a point mass) is: $$a=-{GM\over r^2}$$ The minus sign is because the acceleration is anti-radial. Then you can do the following: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ $$\Delta r~=~\lim_{\Delta t\rightarrow0}-{GM\over r^2}\Delta t^2$$ $$then$$ $$r^2\Delta r~=~-GM\Delta t^2$$ I dropped the limit in the last part because it is implied. If we now use $\lim_{\Delta t\rightarrow0}\Delta t=dt$ and integrate: $$\int_{R_o}^{r_f}r^2dr~=~\iint_0^t-GMd^2t$$ $R_o$ is any initial radius and $r_f$ is any final radius (although because this derivation assumed a zero initial velocity, if $r_f>R_o$ it all breaks down). And after some razzmatazz algebra: $$r_f(t)~=~\sqrt[3]{R_o^3-{3\over2}GMt^2}$$ And if you want to check it, type this into Wolfram, differentiate it twice, plug in the radius of earth, its mass, and $t=0$ and you'll find it says the acceleration is $-9.8{m\over s^2}$