No, because there's no deterministic way to infinitely extend that sequence. In your first example:
1 3 3 3 2 2 2 1 1 1 4 4 x x y y
1 3---- 3---- 3---- 2-- 2-- 2--
>Whenever you start a new count, choose a random number and repeat for a many times as needed for the trailing sequence to match the top sequence.
You answered your own question. The Kolakoski sequence is special because it does not just pick a random number: the sequence is deterministically encoded by the run lengths, and vice versa.
It's not like i had any difficulty coming up with that sequence i wrote to that point.
1 3 3 3 2 2 2 1 1 1 4 4 5 5 6 6 8 1 2... as an example of how trivial it is to continue to this.
I get that if you limit yourself to '1' or '2' you force the choice of next number but even then there's two possibilities of this sequence (start on 1 vs start on 2).
2, 2, 1, 1, 2, 1, 2...
2, 2, 1, 1,
Again no randomness. Just a variant of the above and trivial to continue this.