Could we send a man safely to the Moon in a rocket without knowledge of general relativity? I'm wondering if it's possible to send a man to the Moon using equations consistent with Newtonian gravity and without the elaborate tools of Einstein gravity. Are the predictions made by Newtonian gravity sufficiently precise to plan a successful voyage? If not, where would the Newtonian equations fail and how does Einstein gravity correct for these deficits?
 A: The Jet Propulsion Laboratory has incorporated general relativistic effects in its numerical integration of the planets since the mid to late 1960s. For example, the JPL DE19 ephemeris, released in 1967, incorporated relativistic effects in its modeling of the solar system.
This didn't help much. Had they ignored relativistic effects there would have been little effect. Errors in those older JPL ephemerides piled up quickly, degrading to uselessness in just a few years. Most of these errors were due to extremely lousy computational capabilities (your laptop / home computer is much more powerful than the largest supercomputer of the 1960) and rather lousy measurements (the JPL Deep Space Network was still in its infancy).
Other parts of NASA, including other parts of JPL, did not incorporate relativistic effects into their propagation of their spacecraft. There was little point. Back in the 1960s, NASA's model of the Earth's gravity field was a 4x4 spherical harmonics model, and of the Moon, a simple spherical gravity model. (Compare that to the 2159x2159 EGM2008 Earth gravity model and the 900x900 GRGM900C lunar gravity model.) The errors induced by these known limitations dwarf the error by not modeling those tiny relativistic effects.
In 1968, NASA was quite shocked by the 2 kilometer errors they were seeing in their lunar probes and in the 1968 flight of Apollo 8. This was something NASA did chase down. It turns out that the near side of the Moon has five large mass concentrations that make a mockery out of that simple spherical gravity model. This problem was worth correcting.
Not modeling relativistic effects? In many cases, that's still not worth correcting. Until very recently, I was the key architect of much of the orbital mechanics software used at the NASA Johnson Space Center. I asked on an annual basis to be able to add relativistic effects to our gravitational computations. That request was turned down, every year. I asked because I wanted to put it in, not because it's important for modeling spacecraft behavior.
General relativity has a tiny, tiny effect on spacecraft. They're not up long enough to see the errors that result from ignoring those effects to grow. Ignoring relativistic effects induces a tiny, tiny error in the propagated state, one that is completely swamped by other errors. For example, in the case of a vehicle in low Earth orbit, the uncertainties in the Earth's upper atmosphere are huge. A small solar flare is all it takes to make the Earth's upper atmosphere swell like a balloon. There's no point in modeling relativistic effects when drag is several orders of magnitude higher and when you're lucky to know drag to two places of accuracy.
In the case of a vehicle going to the Moon or to another planet, the errors in the guidance, navigation, and control systems once again swamp the effects of ignoring relativity. These errors, along with others, need to be corrected lest the spacecraft miss the target. Every spacecraft going to some other solar system body needs to make at least one midcourse correction along the way. At worst, ignoring relativistic effects merely means having to bring a tiny bit of extra fuel for those midcourse corrections.
A: Consider that it would not be particularly difficult to do an Apollo-type landing if each of your relative velocity, range, and angular measurements were off by +/- 5%.
You could simply make small iterative corrections along the way, until the absolute values were small enough to make the relative errors inconsequential. At worst you'd need to carry somewhat more safety margin in fuel.
A: A few sanity checks without actually computing anything:
First, the error due to neglecting general relativity is so small that it didn't affect prediction of lunar eclipses and wasn't actually noticed anywhere except in Mercury's orbit (at least not until they purpose-built experiments to detect minor discrepancies). I know this doesn't give a completely satisfying answer, but the moon and our rockets follow the same physical laws and if Kepler mechanics is good enough for the moon, it's good enough for the rocket.
Second, the precision of rocket thrust force and duration, especially half a century ago, was limited. Precisions in brute-machinery engineering (which a solid fuel burning rocket definitely qualifies as) are optimistically around three decimal places, which is much worse than the precision to which the Newtonian dynamics was verified experimentally to hold.
Third, course adjustments are made during flight to correct the trajectory and arrive at the correct destination. So we are not relying on an extremely precise launch to eliminate the accumulation of errors.
In short, general relativity effects are completely overshadowed by imperfections of hard metal machinery, poorly measured amounts of fuel, fuel impurities, combustion irregularities, turbulence and general aerodynamics in the atmosphere during launch, inaccurately determined payload weight, bird poop on the windscreen and so on. Given that we arrived at the moon with steampunk technology, general relativity can be safely ignored for space-travel, at least if you are far enough from a star.
That of course doesn't mean that general relativity isn't noticable elsewhere. GPS is definitely affected by it, and our timekeeping is also accurate enough to detect a difference in time if you climb on a mountain and come back down.
A: I'll start the ball rolling on this one. My GR knowledge is probably not good enough to make this a truly satisfying answer...
The gravitational acceleration for an object moving radially at non-relativistic velocities in the Schwarzschild metric is modified by a factor $(1 - r_s/r)(3[1-r_s/r] -2)$, where $r_s = 2GM/c^2 = 0.00885 m$ for the Earth.
If we take a low-Earth orbit of a few hundred kilometres, the factor is $0.999999995$. For somewhere between the Earth and the moon it is $0.9999999998$.
So if you do something dumb like use a uniform acceleration equation for 3 days, then the (radial) positional inaccuracy that results comes from the gravitational field times $t^2$ multiplied by the factors above. I think the second factor is more realistic since most time was spent between the Earth and the Moon. The gravitational field here is of order 0.02 m/s$^2$, giving an order of magnitude positional error after a 3 day flight of 0.3 metres, or a bit bigger if more time is spent in a stronger gravitational field.
Tangentially I guess we can do an order of magnitude calculation by using the Schwarzschild metric time dilation for the Earth at the moon's orbit. To first order, a clock at the moon runs faster than one at the Earth's surface by a rate of $(1- r_s/r)^{-1/2}$, where $r \sim 6,400\ km$. Multiplying this by 3 days, leads to a temporal error of 0.2 milli-seconds, during which time the moon has moved (wrt Earth) by about 0.2 metres. 
So this extraordinarily rough calculation seems to indicate that GR is nothing to worry about here. But I'm sure someone can do a more accurate job. In any case, I don't think the premise of the question is correct, since in-transit and in-orbit flight corrections could and were done (several times) during the Apollo flights.
A: The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction.
I hope everyone who reads this site will known that gravitational potential energy is given by Newton's law:
$$ V(r) = -\frac{GMm}{r} $$
The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:
$$ V_{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$
where $L$ is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we calculate $V_{eff}$ for an object in a Earth-Moon transfer orbit we get a graph like this:

The stable circular orbit is at the minimum of the potential i.e. at about 384,400km, which is reassuring as this is the Earth-Moon distance. So far so good.
But when we include the effects of General Relativity we find that it modifies the equation for the effective potential. The details are given in the Wikipedia article on Schwarzschild geodesics, but let's skip the details and just give the equation for $V_{eff}$ including relativistic effects:
$$ V_{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3} \tag{2} $$
So including relativistic effects just adds a third term in $r^{-3}$.
Now we calculate the position of the stable orbit by finding the minimum of $V_{eff}$ i.e. we calculate $dV/dr$, set it to zero and solve the resulting equation for $r$. Doing this for the Newtonian potential (1) gives us:
$$ r = \frac{L^2}{GMm^2} \tag{3} $$
Finding the minimum of the relativistic expression (2) is a little more involved as we end up with a quadratic to solve, but some fiddling around ends up with:
$$ r = \frac{L^2}{2GMm^2} \left( 1 + \sqrt{1 - \frac{12G^2M^2m^2}{L^2c^2}} \right) $$
and we can approximate the square root using the binomial theorem to get:
$$\begin{align}
 r &\approx \frac{L^2}{2GMm^2} \left( 1 + 1 - \frac{6G^2M^2m^2}{L^2c^2} \right) \\
   &\approx \frac{L^2}{GMm^2} - \frac{3GM}{c^2} \tag{4}
\end{align}$$
And comparing our calculated Newtonian (3) and relativistic (4) distances we find the difference between them is:
$$ \Delta r \approx \frac{3GM}{c^2} \approx 1.3 \text{cm} $$
So that's how much difference including general relativity makes to the calculated Earth-Moon transfer orbit - about 1.3cm!
A: Absolutely we could, and in fact, I strongly suspect that General Relativity was never used in the Apollo program. for one thing, the on-board navigation computers were nowhere near powerful enough to perform any useful calculation with GR. 
on the other hand, it's possible to measure the position of the moon to within a few centimeters (much more accurate than is necessary to get a spacecraft there safely) and it's certainly necessary to use GR to model that data accurately.
