Does "ZF + no amorphous sets" implies some form of choice?

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If we forbid existence of amorphous sets, then does that prove axiom of choice? Or some form of choice?

Where by a "form of choice" I mean something along those lines.

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No, not at all.

The Cohen model, for example, does not have any amorphous sets. The second Cohen model does not have any either, although the proof is slightly more involved.

In either case, countable choice fails, and there are Dedekind-finite sets. In the latter case, countable choice from families of pairs fails as well.