Yes, the composition factors are unique up to permutation, and this can be derived from the Jordan-Hölder theorem for finite groups.
If $G$ is second countable, it has a composition series: one can obtain this by starting with some countable basis of identity neighbourhoods (which may be taken to consist of open normal subgroups of $G$, since every open neighbourhood of the identity contains an open normal subgroup), taking intersections to form a descending chain of open normal subgroups with trivial intersection, and then refining the series. So let us assume that we have a composition series $(G_n)$. The equivalence class of $(G_n)$ is defined by the number of times, say $f(S)$, each isomorphism type $S$ of finite simple group appears as a factor $G_{n-1}/G_n$.
Let $S$ be a finite simple group. Then there are two possibilities: either $f(S)$ is a non-negative integer, or $f(S) = \aleph_0$. If $f(S)$ is finite, then there is a finite discrete quotient $G/G_n$ of $G$ such that some (hence any) composition series for $G/H$ has $f(S)$ simple factors isomorphic to $S$. Moreover, every finite discrete quotient of $G$ has at most $f(S)$ copies of $S$ in any composition series: This is clear for $G/G_n$ for all $n$ assuming Jordan-Hölder for finite groups, and then we use compactness to show that any finite discrete quotient $G/H$ of $G$ is a quotient of $G/G_n$. If $f(S)$ is infinite, then $G$ has finite quotients with composition series that have arbitrarily many factors isomorphic to $S$. Thus $f(S)$ is an invariant of $G$ as a topological group, independent of the choice of composition series $(G_n)$, so all composition series are equivalent.
One can define transfinite descending composition series for arbitrary profinite groups. Here I am not sure about uniqueness: the difficulty is in counting the cardinality of times a composition factor appears (assuming it appears infinitely many times).