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Jim Conant
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You can cook up lots of normal subgroups by looking at any characteristic subgroup of the free group. For example, if $F^{(k)}$ is the $k$th term of the lower central series, there is a surjection $$Out(F_n)\twoheadrightarrow Out(F_n/F_n^{(k)}).$$ The kernel of this surjection is an interesting normal subgroup. Oscar's construction is a special case of this.

(Edit: I misread "perfect" as "simple," so this is not an answer to the question.)

Jim Conant
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