What is the physical meaning of an LC circuit having a negative transfer function?
Does the signal go 180 degrees out of phase with the input?


Yes it does - that's what the negative implies 

At resonance (\$\omega = \dfrac {1}{\sqrt{LC}}\$) the phase angle passes through -90º from 0º on its way to 180º. 

It's also interesting to note that at resonance the formula becomes infinite in value too because there is no damping in the circuit to restrict it.

How does the voltage across C become infinite? It's because the input impedance to the series combination of L and C (at resonance) is zero and therefore infinite current flows. Totally impractical of course and the more regular analysis is done with a series R added: -

\$\dfrac{Vc}{Vin}\$ = \$\dfrac{\frac{1}{LC}}{s^2 + s\frac{R}{L} + \frac{1}{LC}}\$

Where \$\frac{1}{\sqrt{LC}}\$ is the natural resonant frequency.

This time, at resonance the denominator has \$j\omega \frac{R}{L}\$ left in the formula. Here's what the amplitude and phase response looks like: -



R I've chosen to be 1\$\Omega\$, L is 100uH and C is 25nF. It peaks (because R is quite low) at over 30dB but you can see the phase does all the interesting stuff at close to resonance where it shifts 180º over a small range of frequencies.