There is no Jacobi-like functional equation for your $\phi(x)$. Nevertheless, with a bit of analysis, we can determine its asymptotic behavior as $x\to\infty$ and as $x\to 0+$.

When $x>0$ is large, the leading term $p=2$ dominates, so we have the asymptotics
$$\phi(x)\sim e^{-4\pi x},\qquad x\to\infty.$$

When $x>0$ is small, we use the prime number theorem in the form
$$\pi(x)\sim\mathrm{Li}(x):=\int_2^x\frac{dt}{\log t}.$$
Using this result and integration by parts, we get, as $x\to 0+$,
\begin{align*} \phi(x)&=\int_{2-}^\infty e^{-t^2\pi x}\,d\pi(t)
=\int_2^\infty \pi(t)\,d(-e^{-t^2\pi x})\\
&\sim \int_2^\infty \mathrm{Li}(t)\,d(-e^{-t^2\pi x})
=\int_2^\infty e^{-t^2\pi x}\,d\,\mathrm{Li}(t)
=\int_2^\infty e^{-t^2\pi x}\,\frac{dt}{\log t}.
\end{align*}
Making the change of variable $t=u/\sqrt{x}$ in the last integral, we get
$$\phi(x)\sim\frac{1}{\sqrt{x}}\int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})},\qquad x\to 0+.$$
Now the denominator $\log(u/\sqrt{x})$ always exceeds $2$, but for the bulk of the integral it is $\sim\log(1/\sqrt{x})$. More precisely, with the notation $r(x)=\sqrt{\log(1/x)}$, we can estimate the integral as
\begin{align*}
\int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})}&=\int_{2\sqrt{x}}^{\exp(-r(x))}+\int_{\exp(-r(x))}^{\exp(r(x))}+\int_{\exp(r(x))}^\infty\\[8pt]
&=O(e^{-r(x)})+\frac{1+o(1)}{\log(1/\sqrt{x})}\int_{\exp(-r(x))}^{\exp(r(x))}e^{-u^2\pi}\,du+O(e^{-r(x)})\\[8pt]
&=\frac{1/2+o(1)}{\log(1/\sqrt{x})}=\frac{1+o(1)}{\log(1/x)}.
\end{align*}
Hence we have proved that
$$ \phi(x)\sim\frac{1}{\sqrt{x}\log(1/x)},\qquad x\to 0+.$$