Is the cosmological horizon of an accelerated observer spherical? I guess the question says it all. If you move across the universe (I mean: across the cosmic background radiation) with accelerated motion, does your cosmological horizon (I mean: the one you see) remain spherical - or does it become oblate or prolate?
 A: For our universe the geometry of the horizon for an accelerated observer is a complicated problem, and most importantly depending on the nature of the dark energy, which is unclear for now, so instead, let us consider a simpler question of the horizon of a uniformly accelerated observer in a de Sitter universe (a universe without ordinary matter and with cosmological constant).
Since de Sitter space is maximally symmetric, we can offer a simple intuition for a uniformly accelerating observer in such  spacetime. First, let us pick an inertial (non-accelerating) observer, and rigidly attach to her a second observer held at a fixed distance (and fixed angular variables) from the first one. This second observer would be experiencing constant acceleration toward the first, inertial observer. The situation could be visualized as a first observer dragging the second against the Hubble flow with a rope of fixed length. The horizons seen by both observers are the same, since any signal that could reach one could also reach the other. So a horizon seen by an accelerated observer is simply a spherical cosmological horizon of an inertial observer located forward along the direction of acceleration.
While the internal geometry of the horizon of an accelerated observer is spherical, because the spacetime is curved, its appearance would be distorted, more so for larger accelerations. 

Technical details
The de Sitter universe as seen by an accelerated observer could be best understood with the help of accelerated coordinates, defined for example in the paper:


*

*Podolský, J., & Griffiths, J. B. (2000). Uniformly accelerating black holes in a de Sitter universe. Physical Review D, 63(2), 024006, doi:10.1103/PhysRevD.63.024006, arXiv:gr-qc/0010109.


The metric in this coordinates takes the following form:
$$
d s^2 = {\ 1\over[\sqrt{1+a^2A^2} + A\,r\cos\theta]^2}
 \Bigg\{{d r^2\over1-{r^2\over a^2}} 
 +r^2(d \theta^2+\sin^2\theta\,d\phi^2) -
 {\textstyle \left(1-{r^2\over a^2}\right)d t^2\ }\Bigg\} \>.
$$
Here $a$ is de Sitter length parameter, $A$ is the absolute value of 4-acceleration of observer at $r=0$,  $|A|\equiv |\dot u^\mu |$ (it is identical to the modulus of the 3-acceleration measured in the natural local orthonormal frame of the observer in units with $c=1$). This metric is conformal to the standard static form of the de Sitter metric  to which it reduces when $A= 0$. Cosmological horizon in this coordinates is located at $r=a$, but because the conformal factor depends on the angle $\theta$ there is no spherical symmetry anymore. In the direction of acceleration (it corresponds to $\theta=\pi$) the distance to the horizon is larger than in the non-accelerating case while  in the opposite direction ($\theta=0$) horizon is closer.
For small accelerations, ($aA\ll 1$) the distance between accelerated observer and the corresponding inertial one is approximately $a^2 A$, so the cosmological horizon  is a sphere observed from a slightly off-center position.
For large accelerations $aA \gg 1$ the horizon is a short distance $A^{-1}$  behind the acceleration direction while in the forward direction the distance to the horizon is double the distance from  inertial observer. At a length scales much smaller than $a$ the horizon behind could be interpreted as a planar Rindler horizon.
Note, that if we approximate our universe by a de Sitter space, the accelerations for which $Aa=1$ are very small from the everyday viewpoint, about $10^{-9}\,\text{m/s}^2$.
