smooth functional to detect whether a function has a zero
For the second question only:
For $f(x,t) = (x-t)^2$, you have that if $t > 1$
$$ \int_0^1 \frac{1}{f(x,t)} ~dx = \frac{1}{t-x} \Big|^1_0 = \frac{1}{t - 1} - \frac{1}{t} = \frac{1}{t(t-1)} $$
So you functional $$ G(f)(t) = \begin{cases} 0 & t \leq 1 \\ t^2(t-1)^2 & t > 1\end{cases} $$
and is not a smooth function of $t$ (not even $C^2$).
This example can of course be circumvented if you take instead of $(\int_0^1 \cdot dx)^{-2}$ something like $\exp - (\int_0^1 \cdot dx)$.
A functional that works is $$ F(f)=\begin{cases} \exp(-\exp(\int_0^1 \frac{1}{f(x)}dx)), & \textrm{if $f(x)>0$ for all $x \in [0,1]$} \\ 0, & \textrm{otherwise}. \end{cases} $$ The idea to use this formula and a sketch of the proof that it works are both due to Chengjie Yu. The proof is in the appendix of a paper I wrote with Enxin Wu, Smooth classifying spaces, http://arxiv.org/abs/1709.10517
I don't know whether the functional described in the question, $G(f) = \exp(-\int_0^1 \frac{1}{f(x)} dx)$, works.
Here is the proof that either of the definitions of $G$ in the question satisfy a weaker version of the smoothness condition. I hope that the same ideas can be used to prove that the second definition is in fact smooth. So we'll discuss that case.
Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. We'll show that the first derivative of $G(f)$ with respect to $t$ exists for all $t$. Without loss of generality, we'll do this at $t=0$. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.
If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and we'll show below that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$.
Can the argument be adapted to handle the higher derivatives?
Proof of the inequality (*):
Let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then, by smoothness, there exists $C > 0$ such that $$ f(x,t) \leq C (t^2 + (x - x_0)^2) $$ for every $t \in [-1,1]$ and $x \in [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_x f(x_0,0) = \partial_t f(x_0,0) = 0$. In particular, if $|x - x_0| \leq |t| \leq 1$, then $$ f(x,t) \leq 2 C t^2 . $$ Choose $t \in [-1,1]$ such that $f(x,t)$ has no zeros for $x \in [0,1]$. Integrating, we get $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \int_{\max(x_0 - |t|,0)}^{\min(x_0 + |t|,1)} \frac{1}{f(x,t)} dx \geq \int_{\max(x_0 - |t|,0)}^{\min(x_0 + |t|,1)} \frac{1}{2 C t^2} dx \geq \frac{|t|}{2 C t^2} = \frac{c}{|t|}, $$ as claimed. The last inequality comes from the fact that the interval of integration has width at least $|t|$.
That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation one could simply replace the integrand $1/f$ with $1 / f^2$.