Indefinite double integral

The answer is affirmative. If it is assumed in the sequel that all functions are "neat", then we have: $$ u(p,q) = \iint f(p,q)\, dp\, dq = \iint f(p,q)\, dq\, dp \\ \Longleftrightarrow \quad \frac{\partial^2}{\partial q \, \partial p} u(p,q) = \frac{\partial^2}{\partial p \, \partial q} u(p,q) = f(p,q) $$ In particular, if the cross partial derivatives are zero: $$ \frac{\partial^2}{\partial q \, \partial p} u(p,q) = \frac{\partial^2}{\partial p \, \partial q} u(p,q) = 0 $$ Do the integration: $$ u(p,q) = \iint 0 \, dq\, dp = \int \left[ \int 0 \, dq \right] dp = \int f(p) \, dp = F(p) $$ On the other hand: $$ u(p,q) = \iint 0 \, dp\, dq = \int \left[ \int 0 \, dp \right] dq = \int g(q) \, dq = G(q) $$ Because $\;\partial f(p)/\partial q = \partial g(q)/\partial p = 0\,$ : that's the meaning of "independent variables".
We conclude that the general solution of the PDE $\;\partial^2 u/\partial p \partial q = \partial^2 u/\partial q \partial p = 0\;$ is given by: $$ u(p,q) = F(p) + G(q) $$ This result is more interesting than it might seem at first sight.

Lemma. Let $a\ne 0$ and $b\ne 0$ be constants (complex eventually) , then: $$ \frac{\partial}{\partial (ax+by)} = \frac{1}{a}\frac{\partial}{\partial x} + \frac{1}{b}\frac{\partial}{\partial y} = \frac{\partial}{\partial ax} + \frac{\partial}{\partial by} $$ Proof with a well known chain rule for partial derivatives (for every $u$): $$ \frac{\partial u}{\partial (ax+by)} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial (ax+by)} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial (ax+by)} $$ Where: $$ \frac{\partial x}{\partial (ax+by)} = \frac{1}{\partial (ax+by)/\partial x} = \frac{1}{a} \\ \frac{\partial y}{\partial (ax+by)} = \frac{1}{\partial (ax+by)/\partial y} = \frac{1}{b} $$ Now consider the following partial differential equation (wave equation): $$ \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0 $$ With a little bit of Operator Calculus, decompose into factors: $$ \left[ \frac{\partial}{\partial c t} - \frac{\partial}{\partial x} \right] \left[ \frac{\partial}{\partial c t} + \frac{\partial}{\partial x} \right] u = \left[ \frac{\partial}{\partial c t} + \frac{\partial}{\partial x} \right] \left[ \frac{\partial}{\partial c t} - \frac{\partial}{\partial x} \right] u = 0 $$ With the above lemma, this is converted to: $$ \frac{\partial}{\partial (x-ct)}\frac{\partial}{\partial (x+ct)} u = \frac{\partial}{\partial (x+ct)}\frac{\partial}{\partial (x-ct)} u = 0 $$ With $p = (x-ct)$ and $q = (x+ct)$ as new independent variables. Now do the integration and find that the general solution of the wave equation is given by: $$ u(x,t) = F(p) + G(q) = F(x-ct) + G(x+ct) $$ Interpreted as the superposition of a wave travelling forward and a wave travelling backward.

Very much the same can be done for the 2-D Laplace equation: $$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $$ Decompose into factors (and beware of complex solutions): $$ \left[ \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right] \left[ \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right] u = \left[ \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right] \left[ \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right] u = 0 $$ This is converted to: $$ \frac{\partial}{\partial (x+iy)}\frac{\partial}{\partial (x-iy)} u = \frac{\partial}{\partial (x-iy)}\frac{\partial}{\partial (x+iy)} u = 0 $$ With $\;z=x+iy\;$ and $\;\overline{z}=x-iy\;$ as new, complex, independent variables.
Now do the integration: $$ u(x,y) = F(z) + G(\overline{z}) $$ The solutions are related to holomorphic functions in the complex plane.