On a formal level, the functional integral is an infinite-dimensional generalization of a scaled Lebesgue integral in $R^n$. For the latter, any diffeomorphism (bijective $C^1$ transformation) of the vector $x$ over which the integration extends can be compensated by changing the measure by the corresponding absolute value of the Jacobian determinant. In infinite dimensions, $x$ becomes a field, but the transformation law should still be valid.
However, since there is no proper definition of the functional integral (and an infinite-dimensional Lebesue integral doesn't exist) it is not clear to which extent the above reasoning is valid. People use it (just like the functional integral itself) on a heuristic basis.
Note that the property of renormalizability is not preserved under nonlinear field transformations, but renormalizability is needed to give the functional integral a well-defined and unique perturbative meaning. Thus performing a nonlinear tranformation of fields in the context of functional integrals is a hazard that easily leads to inconsistent results.
A proper answer to your question must wait for the day where functional integrals have a meaningful nonperturbative definition.