Is the bitonic sort algorithm stable?
No, bitonic sort is not stable.

For this post I will denote numbers as 2;0 where only the part before the ; is used for comparison and the part behind ; to mark the initial position.
Comparison-exchanges are denoted by arrows where the head points at the desired location of the greater value.

As written in the link that @JukkaSuomela posted a stable sorting network needs to avoid swaps of equal values.

When swapping equal values, the bitonic sorter for two values is already unstable:

0;0 ----- 0;1
      |
      v
0;1 ----- 0;0


Of course, this can be fixed when we don't swap equal values:

0;0 ----- 0;0
      |
      v
0;1 ----- 0;1


However, it could happen that the order of two equal elements is swapped without them being compared to each other.

This is exactly the case in this example of a bitonic sorter for 4 values:

1;0 ------ 1;0 ------ 0;2 ------ 0;2
      ^          |          |
      |          |          v
1;1 ------ 1;1 --|--- 1;1 ------ 1;1
                 ||
                 v|
0;2 ------ 0;2 ---|-- 1;0 ------ 1;0
      |           |         |
      v           v         v
2;3 ------ 2;3 ------ 2;3 ------ 2;3


Although we were careful not to swap elements that compared equal (upper left comparison), the merging pass swapped the order of 1;1 and 1;0 which cannot be corrected later on.

This counterexample proves that bitonic sort cannot be stable.