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] 1 results from [the difference in accelerations due to the difference in gravitational potentials] 2 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, one needs to remember that SR assumes inertial frames. There are claims, however, that it can be shown that accelerations are easily handled by SR. Well, I have shown in [my answer here] 3, how it is being achieved through mere sleigh of hand.
So, it is not that SR claims certain things about acceleration. It is simply certain people who claim these things.