By replacing $\phi$ by $-\phi$ you can set $\alpha$ to be positive (or negative, if you wish). By replacing $\phi$ by $\lambda \phi$ you can rescale away $\alpha$. So you can set $\alpha$ to be either $+1$ or $-1$ as you wish.

So for now let's consider the situation you are solving
$$ \partial^2_{tt}\phi - \Delta \phi = |\nabla \phi|^2 $$
First observation: suppose $\phi$ is a $C^3$ solution with compactly supported initial data, then $\phi$ is uniformly bounded from below on $\mathbb{R}^3\times [0,\infty)$. This is due to the fact that on $\mathbb{R}^3$ the fundamental solution to the wave equation is signed (a fact also used by John in his proof), so that the solution to the above equation is pointwise $\geq$ the solution to the linear equation with the same initial data. The latter is globally uniformly bounded, and hence $\phi$ is uniformly bounded below.

Now $v = e^\phi - 1$; you have that $v$ is also $C^3$, and there exists $\beta$ such that $v + 1 > \beta$ uniformly on $\mathbb{R}^3\times [0\infty)$, using the uniform bound on $\phi$.

Now we compute
$$ (\partial^2_{tt} - \Delta) v = e^\phi (\partial^2_{tt} - \Delta) \phi + e^\phi (\partial^2_t \phi - |\nabla\phi|^2) = \frac{1}{1+v}( \partial_t v)^2 $$

The lower bound above ensures that the denominator on the RHS is never zero.

Suppose now that $\phi$ is globally bounded *above* also: then $(1+v)$ is bounded above globally by some $M > 0$. Then the condition for Theorem 2 in John's 1981 paper is satisfied and the result follows.

So the remaining question is whether it is possible for $\phi$ to blow-up only at infinity and not at any finite time. I think there may be a way to rule this out, but that would require checking some more technical details so I'll have to report back later.