Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of
$$F = \frac{G m_1 m_2}{r^2}$$
where $r = x_2 - x_1$ (assuming $x_2 > x_1$) is the separation distance. So for object 1, you have
$$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$
and for object 2,
$$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$
Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get
$$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$
So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get
$$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$
and integrate,
$$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$
Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to
$$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$
where I've chosen the negative square root because $\dot{r} < 0$, and integrate it again to find
$$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} + r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$
where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time.
In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series,
$$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$
The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.
This is a tricky question. Fundamentally, this is the motivation of general relativity (and all metric theories of gravity)--if all masses interact with a gravitational field in the same way, then, in a sense, the motion of a particular mass is determined by the local gravitational field, independently of the mass. This then leads you into explaining the gravitational "force" as an emergent property of the local spacetime curvature.
But then, what came first? The explanation of gravity as curvature, or the equivalence of gravitational and inertial mass? In a way, they are just dual pictures of the same thing.
Best Answer
As Dale points out there are tests of the weak equivalence principle that check that the acceleration of bodies of different composition, towards a large mass, are the same.
That checks that the 'passive gravitational mass' is equivalent to inertial mass. The experiments show that the equivalence principle is true to a high degree of accuracy.
It means that one aspect of Einstein's General Relativity seems to be valid. The motion of the masses can be modelled by a 'bending of space-time' by the large mass.
If you mean, in the question - Is there a way to check that the 'active gravitational mass' is equivalent to the inertial mass? Then it can't easily be accurately checked.
(the terms are explained here in the section weak equivalence principle)
It would mean checking that the amount that a mass bent space-time is proportional to it's inertial mass. As you say we can't know the inertial mass from the passive gravitational mass, so another means (electrostatic) would be needed to determine it .
That seems to be limit the objects mass. Then it seems that, even if the gravitational attraction from that mass could be measured, the accuracy would be limited by our knowledge of $G$, unfortunately $G$ is only known to about $0.6$%.
It could well be that this type of experiment has never been carried out, there are likely to be large uncertainties. It would be interesting to know to what extent the active gravitational mass has been found to be equivalent to inertial mass.
Perhaps an expert in General Relativity will comment whether the equivalence follows once GR is accepted, although it would be a theoretical justification and not an answer to your 'Has anyone directly observed...' question.