I'm going to make some educated guesses and grossly oversimplify the problem:
First, we'll ignore the change in gravitational acceleration over the 200km fall (it's only a 5% difference, so who cares). Next we'll assume that our spacecraft encounters negligible atmospheric resistance until it reaches the Karman line at 100km. Finally we'll assume that it begins to decelerate once it has reached the Karman line.
After a 100km fall the speed of the craft can be found with the equations of motion thus:
$$s=ut+\frac{at^2}{2}$$
Taking $$g = 9.8 ms^{-2}, u = 0 ms^{-1}, s = 100 km$$ and solving for t gives: $$t = 140s$$
Putting this value into:
$$v = u + at$$
and solving for v we get: $$v = 1400 m/s$$
Which is only 3100 mph, or Mach 4 (at sea level). For comparison, orbital velocity starts around 16,800 mph (Mach 22).
So from this calculation we can state that no, provided the craft is reasonably aerodynamic it doesn't need a heat shield to survive reentry. The craft would get a little warm, but it would be possible to work with it. Once it lost enough energy to go sub-sonic it would be easy enough to bring the craft down on parachutes.
However, even if our assumption that the spacecraft would start to decelerate at the Karman line was wrong, and the craft didn't start slowing until 0 altitude (maybe they where aiming at the Dead Sea), the craft would still only be going at 4400 mph (Mach 6). With a little cooling, some thermal paint, and smooth lines we'd still be fine without a heat shield. (Although faced with the bigger problem of surviving a 4400 mph collision with a medium sized planet)
A real world example of this kind of sub-orbital reentry can be found in sounding rockets. These craft routinely reenter the atmosphere from apogees of 1,000km to 1,500km and generally suffer greater heating on the way up than they do coming back down. Most don't have a dedicated heat shield, just thermal paint.