Definition of Force in Lagrangian Mechanics

Typically, the Newtonian definition of force is given in terms of the rate of change of momentum,
$frac{dvec{p}}{dt}$.
But in trying to connect to the classical Lagrangian and Hamiltonian formulations of mechanics, one often comes up against canonical momenta which do not perfectly match the familiar momentum of a free particle,
$vec{p} = m dot{x}$.
An example is the Lagrangian for a charged particle in an electromagnetic field,
$$L = frac{m}{2} dot{x}^2 + q vec{A} cdot dot{x} – qphi$$
where $phi$ and $vec{A}$ are the electromagnetic scalar and vector potentials, respectively. The canonical momentum for this Lagrangian is,
$$vec{p} = frac{partial L}{partial dot{x}} = m dot{x} + qvec{A} $$
After solving the Euler-Lagrange equations for this Lagrangian, one obtains,
$$m ddot{x} = q vec{E} + q dot{x} times vec{B}$$
which is typically defined to be the Lorentz force. But this force is decidedly NOT the same as the time derivative of the canonical momentum associated to this Lagrangian, but rather the time derivative of the "kinetic momentum,"
$$vec{P} = vec{p} – qvec{A} = m dot{x}$$
An almost identical problem comes up when using the special relativistic Lagrangian for a charged particle in a field,
$$- mc^2 sqrt{ 1 – left( frac{dot{x}}{c} right) ^2 } + qvec{A} cdot dot{x} – qphi $$
whose canonical momentum is,
$$vec{p} = frac{m dot{x}}{sqrt{1 – left( frac{dot{x}}{c} right) ^2 }} + q vec{A} equiv gamma m dot{x} + q vec{A}$$
After solving the Euler-Lagrange equations again, we obtain,
$$frac{d}{dt} [ gamma m dot{x} ] = q vec{E} + q dot{x} times vec{B}$$
which just looks like the time derivative of the familiar free particle relativistic momentum,
$$vec{P} = vec{p} – qvec{A} = gamma m dot{x}$$
rather than the time derivative of the canonical momentum. Is this just the definition of force in Lagrangian and Hamiltonian mechanics, or is there some deeper reason that the force does not involve the canonical momentum? I have read the (unjustified) assertion that only the kinetic momentum is "measurable," but the meaning of this is unclear, and it doesn’t seem to help resolve this issue.