No.  The set of primes can be whatever you want (added: within reason!  As Benjamin Steinberg points out, it can in fact be any *recursively enumerable* set of primes).

First, note that for infinitely presented groups, the torsion can be whatever you like: the torsion in the group   $*_i \mathbb{Z}/p_i$  is precisely the set of primes $p_i$, by standard facts about free products.  (Added: As long as the set of $p_i$ is recursively enumerable, then this group admits a recursive presentation on a countable set of generators.)

By Higman's Embedding Theorem, the above group can be embedded in a finitely presented group.  More subtlely, this embedding doesn't introduce any new torsion---see, for instance, Theorem 2.5 of [this preprint of Chiodo][1].

**Clarification:** Higman's Embedding Theorem is commonly stated as only applying to finitely generated countable groups.  In fact, an old construction of Higman, Neumann and Neumann shows how to embed a countably generated group into a 2-generated group; if the countably generated group is recursively presented, then the 2-generated group can be taken to be recursively presented as well.

**Further update:**

Following Benjamin Steinberg's comments below and an argument in [another paper of Chiodo][2], I think we have a *very exciting* characterization of the sets of primes that can occur as torsion in finitely presented groups.  (For the solvable-word-problem case, one also needs a theorem of Clapham, which says that the Higman Embedding can be made to preserve solvability of the word problem.)

**Very exciting theorem:** Let $P$ be a set of primes.

1. $P$ occurs as the torsion in some finitely presented group if and only if $P$ is an intersection of a recursively enumerable set and a co-re set.
2. $P$ occurs as the torsion in some finitely presented group with solvable word problem if and only if $P$ is recursively enumerable. 

[1]: http://arxiv.org/abs/1107.1489   [2]: http://www.ams.org/mathscinet/search/publdoc.html?arg3=2011&co4=AND&co5=AND&co6=AND&co7=AND&dr=pubyear&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=chiodo&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=2774658