To me, it seems that the answer is no.
Everett's argumentation seems to be flawed by a well-disguised circularity:
In his original paper Rev. Mod. Phys. 29 (1957), 454-462, Everett claims that his theory is one without the need for state reduction (what he calls 'process 1'). But when he discusses observation, he brings in at the very beginning an apparently innocent additional concept, that of a 'good observation'. The formal definition of the latter is the sentence around (10)-(11) of his paper.
According to Everett, a 'good observation' is an interaction that transforms each state $\phi_S \otimes\psi_O$ such that $\phi_S$ is a fixed eigenstate of the measured variable with eigenvalue alpha into a state $\phi_S \otimes\psi_O(\alpha)$ where $\psi_O(\alpha)$ belongs to a set $X(\alpha)$ of states which belong to the awareness of $\alpha$. In particular, this entails that for different $\alpha$, the sets $X(\alpha)$ must be disjoint. Since Everett only allows the unitary dynamics (his 'process 2', see first line of his Section 3), any interaction 'in a specified period of time', must result in a unitary mapping $U$ on the state space of system plus observer.
Therefore, $U$ corresponds a 'good observation' (of the system in state $\Phi_S$) iff there is a self-mapping $\psi_O \to\psi_O(\alpha)$ of the observer state space $X$ such that
$$ U(\phi_S \otimes\psi_O) = \phi_S \otimes\psi_O(\alpha) \forall \psi_O \in X. $$
From his definition, we also see that the mapping $\psi_O \to\psi_O(\alpha)$ maps $X(\alpha)$ into itself, and the whole observer state space $X$ into $X(\alpha)$. Since the interaction is unitary, it is invertible; but the restriction of $U$ to $\phi_S \otimes X $can be invertible only if $X(\alpha)=X$. But this means that there is only a single eigenvalue $\alpha$.
Therefore, under the assumptions made by Everett, there are no 'good observations', and since his analysis of the observational process depends on the latter, it is void of any meaning. Indeed, looking closer at the concept of a 'good observation', one can see that it is the projection postulate in disguise. (The above argument loses its power once one allows $U$ to be nonunitary.)
Thus Everett's analysis simply derives the projection postulate by having assumed it, without any discussion, in disguise.