If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. This is calculated by multiplying the number density of the neutrons $n_n$ by, $3p_{f}^2/(10m_n)$, the average KE per fermion in a non-relativistically degenerate gas and where $p_f =(3h^3n_n/8\pi)^{1/3}$ is the Fermi momentum.
So even in a teaspoonful (say 5 ml), there is $1.5\times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a billion or so atom bombs) and this will be released instantaneously.
The energy is in the form of around $10^{38}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about 250 million tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 40km radius near-earth asteroid hitting the Earth at 30 km/s.
So, falling through the Earth is not the issue - vapourising a significant chunk of it is.
Note that the beta decay of the free neutrons that dominate the neutron material is also energetic, but a slow process. On these 10 minute timescales, the neutrons could have exploded to a radius of a tenth of an au.