Does zero free current entail zero $\vec H$? There are two kinds of magnetic fields (different authors give them different names), $\vec B $ and $\vec H$ which are related by the equation $$ \vec B = \mu_o (\vec H + \vec M)$$ where $\vec M$ is the magnetization.
Ampere's law for free currents states $$\oint_C \vec H \cdot d\vec l = I_{free} $$
This is my question: does zero free current entail zero $\vec H$? 
My argument for that is this: since the contour integral of $\vec H$ is zero for all arbitrary curves C in a region of zero free current, $\vec H$ is necessarily zero. However this may not be a mathematically correct argument...
 A: I think your argument is completely flawed.
Consider a uniform H-field. The closed line integral of this field around any loop is zero - and there must be no free current through the loop. Hence there is no free current, yet the H-field is non-zero.
You might be better off thinking about this in terms of free current density.
In this case we can say
$$\nabla \times {\bf H} = {\bf J},$$
so that a zero free current density just means that the H-field has zero curl at that point, nothing more than that.
A: Yes, I agree with your argument.
Don’t forget $\vec H$ and $\mathrm d\vec l$ are vectors!
$\vec H\cdot\mathrm d\vec l$ is a "dot product",
so the $\oint$ sum the same direction of $\vec H\cdot\mathrm d\vec l$,
Which means if $\vec H$ is a constant
you can take it out the $\int$
and it becomes $\vec H \oint\mathrm d l$ (here $\mathrm dl$ is scalar since they are in the same direction).
So if the free current is zero,
$\vec H$ must be zero!
A: For example, if only permanent magnets exist, the free current ($\bf{J}_{\text{free}}$) is zero, but both $\bf{B}$ and $\bf{H}$ are not zero.
