There is a basic difference between "all hilbertian are complemented" and "all projections are uniformly bounded". Here is an example: Let $X$ be a separable Banach space such that every operator from $\ell_2$ into $X$ is compact. Consider $\ell_2\oplus X$ and let $Z\subset \ell_2\oplus X$ be Hilbertian. Then $P_X|Z$ is compact so there exists a decomposition $Z=Z_1\oplus Z_2$ such that $\|P_X|Z_1\|<\epsilon$ and $Z_2$ finite dimensional. Note that norms in this $\oplus$ depend only on Banach-Mazur distance from $Z$ to Hilbert not on $\epsilon$. It suffices to show that $Z_1$ is complemented. For $z\in Z_2$ we have $\|P_{\ell_2}(z)\|\geq c\|z\|$ so $V=:P_{\ell_2}(Z_2)$ is a closed subspace in $\ell_2$, let $Q$ be orthogonal projection from $\ell_2$ onto $V$. Let us define $S(\xi,x)=(P_{\ell_2}|Z_2)^{-1} Q(\xi)$. It is onto $Z_2$ and for for $z\in Z_2$ we get $\|z-S(z)\|$ is small--so $Z_2$ is complemented.As $X$ we can take $\ell_p$ with $1\leq p<2$ but also ANY sum $(\sum_n W_n)_p$ of finite dimensional spaces $W_n$.This shows that no local conditions like type, K-convex etc work without the uniform bound.
Sorry; it is really a comment but I got the space limit so put it here.