Consider the Fourier transform of a multivariable probability density function $Pr(\{x_n\})$, i.e. its characteristic function:
$\int Pr(\{x_n\})e^{-i2\pi\sum\limits_{n}f_nx_n}\prod\limits_{n}{dx_n}$,
which can be considered a Fourier transform with respect to a vector variable $\vec{x}$.
Now suppose instead, we consider a probability density function defined over the space of all functions of t over the region from t = 0 to t = t, $Pr[x(t)]$. We can define a path integral as follows:
$\int Pr[x(t)]e^{-i2\pi\int\limits_{0}^tL(x,\dot{x})dt}D[x(t)]$.
If $Pr[x(t)]$ is constant across all functions, then the integral is just the usual Feynman path integral. In analogy to the above, would it be possible to consider this a Fourier transform with respect to a function, i.e. the characteristic function of $Pr[x(t)]$?
So, the main problem here is a QM amplitude (as one derives from the path integral) is a complex number, while a Fourier transform outputs a function. But nonetheless, let's see what analogy can be drawn:
We'll assume without loss of generality that we're working with one particle. So, in classical mechanics, the Lagrangian, $L(x,\dot{x})$ may be related to the Hamiltonian via Legendre transform, i.e. $L(x,\dot{x}) = p\dot{x} - H(x,p)$, where p is the momentum and is now considered to be independent of x.
Substituting into the formula for the path integral I provided in the OP, we get:
$\int Pr[x(t)]e^{-i2\pi\int\limits_{0}^t(p\dot{x} - H)dt}D[x(t)]$,
which may be rewritten as:
$\int Pr[x(t)]e^{-i2\pi\int\limits_{0}^tpdx - Hdt}D[x(t)]$.
If p and H were considered to be totally independent variables of x(t), then yes, you could consider something like this as a characteristic function on path space. The problem is, they both vary along each path, and the end result of the integral is a complex number. So the answer to the question as written is "no".
Perhaps someone else has additional insight that I don't?