Which clock is the fastest inside an accelerating body? The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance. 
Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed? 

 A: Contrary to one of the answers given, if the distance between the clocks, as observed be either clock, remains constant, the two clocks cannot have the same proper acceleration (the acceleration according to an accelerometer attached to the clock); the clock 'in back' will have greater proper acceleration and, thus, will run slower than the clock 'in front'.  This is a well known and uncontroversial result.  See, for example, Rindler coordinates
If both clocks have the same proper acceleration, the clocks will run at the same rate but the distance between the clocks, according to either clock, will increase; the rocket would necessarily have to stretch.
From the above linked article:

It follows that if a rod is accelerated by some external force applied
  anywhere along its length, the elements of matter in various different
  places in the rod cannot all feel the same magnitude of acceleration
  if the rod is not to extend without bound and ultimately break.

Again, as far as I know, this is uncontroversial and quite easy to show.
As usual, a spacetime diagram is helpful:

Plotted are portions of the worldlines of two clocks with uniform proper acceleration to the right.
When the coordinate time is zero, both clocks are (momentarily) at rest in this frame and both clocks are (momentarily) synchronized with the coordinate time, i.e, both accelerated clocks read zero when the coordinate time is zero.
Now, according to either clock, the distance between the two clocks is constant (both clocks have constant Rindler spatial coordinate).
But, the accelerometer on clock A measures a greater acceleration than the accelerometer on clock B.  This is clearly evident given that the curvature of clock A's worldline is greater.
And, indeed, we see that clock A is running more slowly than clock B.
To summarize these results as it relates to the OP's question, implicit in the question is the assumption that, according to either clock within the rocket, the distance between the clocks is constant.
What has been shown is that
(1) if the distance between the clocks, according to either clock, is constant 
(2) and if the clocks are uniformly accelerating according to accelerometers attached to each clock
(3) then, the two accelerometers must read different accelerations.
This is a straightforward result from SR.  Whether this is, in fact, the correct description of world is, of course, a matter of experiment.  But, what SR predicts is unambiguous.

One final note:  once acceleration is introduced into SR, one must be particularly careful about the concepts of acceleration and distance.
For example, there is proper acceleration, an invariant acceleration, and coordinate acceleration which is frame dependent.  While there can be uniform proper acceleration, uniform coordinate acceleration is impossible.
Further, there are different notions of distance between observers with uniform proper acceleration.  There are, for example, the notions of ruler distance and radar distance.
So, when thinking clearly about an acceleration thought experiment in SR, one must be careful to unambiguously specify the problem.
For example, if one says that two clocks have the same acceleration, it isn't clear if one means the same proper acceleration or the same coordinate acceleration.  The answer one gets from SR crucially depends on the difference.
A: The bottom clock will run slower than the top clock.
The simple way to explain this is to appeal to the equivalence principle. This tells us that locally an acceleration is equivalent to a gravitational field. So if the rocket is accelerating at some acceleration $a$ this is equivalent to two clocks sitting stationary in a gravitational acceleration $a$.
We can use this to calculate the gravitational time dilation between the clocks because to a first approximation the relative time dilation is given by:
$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2 \Delta\Phi}{c^2}}} $$
where $\Delta t_{top}$ is the time interval measured by the top clock, $\Delta t_{bottom}$ is the time interval measured by the bottom clock and $\Delta\Phi$ is the difference in the Newtonian gravitational potential. If the distance between the clocks is $h$, then the difference in the potential is simply:
$$ \Delta\Phi = ah $$
so:
$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2ah}{c^2}}} $$
Let's do this calculation for an acceleration of $1g$ and a rocket length of $100$ m. We're taking the upward direction as positive, which means the acceleration is negative because it points down. The relative time is:
$$\begin{align}
 \frac{\Delta t_{top}}{\Delta t_{bottom}} &= \frac{1}{\sqrt{ 1 + \frac{2 \times -9.81 \times 100}{c^2}}} \\
 &= 1.00000000000001
\end{align}$$
The ratio is possibly better written as $1 + 10^{-14}$ i.e. there are thirteen zeros after the decimal point. This is an extraordinarily small effect, but it can be measured. Indeed it was measured by the Pound-Rebka experiment.
A: The clock ahead is behind in time by an amount depending on the speed of the inertial frame of reference theyare in. If the acceleration is thought of as moving through faster and faster inertial frames of reference then the amount that the clock ahead in distance is behind in time increases as the speed of the inertial frames increases. 
A: If you consider your rocket a rigid body*, than at all times all points on and in it will experience exactly the same acceleration. Therefore there is no difference between the two clocks that might be the source of the difference of the rates of these clocks.
There is no time dilatation here due to equivalence principle saying that acceleration is equivalent to acceleration, since both clocks experience exactly the same acceleration. Gravitational time dilatation results from the difference in accelerations due to the difference in gravitational potentials between clocks' locations. The distance separation between the two clocks in the spaceship does not entail any difference in accelerations due to different distance to the "center of acceleration". They can be treated just like two clocks distance-separated on Earth but located at exactly the same altitude, and therefore experiencing exactly the same gravitational acceleration $g$ (which differs depending on altitude). Therefore the rates of the clocks will be exactly the same.
*I said "if ...", but Alfred Centauri opposed in his comment that nothing can be transmitted instantaneously, so there must be some differences in accelerations. Well ... yes, true... But then, let's think about it for a moment ...
Say, we are applying acceleration $a$ to the lowermost part of the rocket. If the spaceship is not perfectly rigid, the acceleration will be increasing (for a brief moment, or less) throughout its body, but finally the uppermost part of it will achieve the initial acceleration $a$. From this moment on the atoms of the whole body will be transmitting the initial value of the acceleration all the way to the top. One can say that this transmission will be "late", i.e. at any given instant atoms closer to the top will be transmitting the acceleration that the atoms closer to the bottom have experienced a fraction of a second earlier. Sure, but still, it does not matter at all, because the value of the acceleration will be uniform throughout as long as the acceleration applied at the bottom is constant. "Late" acceleration $a$ is still acceleration $a$. Therefore - safe for a brief moment until the whole rocket receives the acceleration $a$ - the whole rocket is experiencing exactly the same acceleration. (So we are back where we were at the beginning - no time dilatation between the clocks in the rocket.)
One last note: Should someone claim there are different phenomena due to acceleration in SR, and therefore accelerations throughout the spaceship do differ, one needs to remember that SR assumes inertial frames. There are claims, however, that it can be proven that accelerations are easily handled by SR. Well, I have shown in my answer here, how it is being achieved through mere sleigh of hand.
Should somebody still feel unconvinced, I will remind you, that SR says that within the moving frame no changes due to speed are showing. That all “strange” phenomena are visible by the outside observer only – through comparison. And here I'm going to hear the objection that this claim pertains to inertial frames only. Sure it does – and so does the whole Special Relativity. Which takes us back to the link above where I showed that one cannot derive equations for SR only to discard the axioms these equations are rooted in.
So, it is not that SR claims certain things about acceleration. It is simply certain people who claim SR claims these things.
