Semiproperness of Namba forcing is indeed equivalent to SCC. Here by SCC I mean the version which appears in Chapter XII, Theorem 2.5 part (2) of Shelah's book: for all large $\theta$ and all wellorders $w$ on $H_\theta$ and all countable $N \prec (H_\theta,\in,w)$ and all $\alpha < \omega_2$, there is an $N' \sqsupset N$ such that $N' \prec (H_\theta,\in,w)$ and $\alpha \le \text{sup}(N' \cap \omega_2)$.
The forward direction is in Shelah's book, and the backward direction appears in Section 3 of Doebler's "Rado's Conjecture implies that all stationary set preserving forcings are semiproper", and is attributed to "folklore". Since the Doebler article doesn't appear on arxiv, and since the version of SCC he uses is slightly different from above, I sketch his short proof here. Assume SCC as in the first paragraph. We show that II has a (very easy) winning strategy in the game that Goldstern described in the comment above. As the game progresses, Player II will construct a $\subset$-increasing sequence $X_n$ of countable elementary submodels of $(H_\theta,\in,w)$, and at the $n$-th move player II simply plays the ordinal $X_n \cap \omega_1$. The model $X_{n+1}$ is chosen so that $F_{n+1} \in X_{n+1}$ and $X_n \subset X_{n+1}$. Let $X_\omega$ be the union of the $X_n$'s and $\delta_\omega = X_\omega \cap \omega_1$. Suppose for a contradiction that II loses the game; then there is some $\alpha_0 < \omega_2$ such that for all $\beta \in [\alpha_0, \omega_2)$ there is an $n_\beta \in \omega$ such that $F_{n_\beta}(\beta) \ge \delta_\omega$. By SCC there is a $Y \sqsupset X_\omega$ with $\alpha_0 \le \text{sup}(Y \cap \omega_2)$ and $Y \prec (H_\theta,\in,w)$. Pick any $\beta \in Y \cap [\alpha_0, \omega_2)$. Since $F_{n_\beta} \in X_{n_\beta} \subset Y$ then $F_{n_\beta}(\beta) \in Y \cap \omega_1 = \delta_\omega$. Contradiction.
Regarding the large cardinal strength: The Sharpe-Welch paper mentions an unpublished proof of Magidor that semiproperness of Namba forcing is equiconsistent with a measurable cardinal. However I don't know the proof, and haven't seen it.