If we suppose that the function is smooth for some time, and the second derivative converges almost everywhere to something as we approach a possible blowup time, then I think we can use the assumptions to show that the second derivative converges in $L^1$; and this excludes the possibility of the most naive type of blowup (that is, a "reasonable" way for the second derivative to converge to something with a Dirac mass somewhere). But I'm not sure how to go further than this.

oops... edited!

In the special case of $\phi(z) = z^2/2$, and restricting to $\lambda < 1$ for definiteness, the function we want to see is convex is proportional to $-\frac{\log(1-\lambda)}{\lambda}$. This function is indeed convex, but it does not strike me as something obvious!

I confirm what jjcale said. Recall that the statement is false if your replace the Gaussian measure by an arbitrary probability measure.