What is the meaning of $U''(x)=0$? Most potentials with a minimum  can be described approximately as a harmonic oscillator. 
So the procedure is to Taylor expand $U(x)$:
$$U(x)=U(0)+U'(0)x+\frac{1}{2}U''(0)x^2 +...$$
If we suppose that the potential is cero at the origin an it has a minimum there, we get:
$$U(x)=\frac{1}{2}U''(0)x^2$$
We take $U''(0)$ to be the spring constant $k$. So the angular frecuency is given by:  $\omega=\sqrt{\frac{k}{m}}$
But what if $U''(0)=0$ and there is still a minumum at zero, like a potential $U(x)=x^4$?
In this case if you blindly apply the formula you get zero frecuency, which is false. Does it just mean that to a small approximation a body will not oscillate?
 A: 
Does it just mean that to a small approximation a body will not
  oscillate?

It means that you must always remember the context in which a formula is valid and not "blindly" apply it.
Where does the formula come from?  Consider the homogeneous differential equation for the harmonic oscillator:
$$\ddot x + \dfrac{k}{m}x = 0$$
with solutions
$$x(t) = Ae^{i\omega t} + Be^{-i\omega t}$$
where
$$\omega = \sqrt{\dfrac{k}{m}} $$
But, for a quartic potential, the force on the mass is
$$F = -k'x^3 $$
thus, the differential equation is non-linear:
$$\ddot x + \dfrac{k'}{m}x^3 = 0 $$
and so one should not expect the motion to be a pure (single frequency) sinusoid.
And, since there is no linear term in $x$, there is no linear approximation and thus no context in which to apply the frequency formula for the harmonic oscillator.
A: As lionelbrits also mentions in his answer, one cannot apply the linearized theory (i.e. the harmonic oscillator approximation) if the leading Taylor coefficients of the potential $U$ vanishes.  [We assume that $x=0$ is still a minimum point for an even potential $U(x)=U(-x)$.] 
In an anharmonic oscillator, the period $T$ will in general depend on the amplitude $A$. However, the period can still in principle be determined from the integral
$$ T~=~ 4  \int_0^A\!dx \sqrt{\frac{m/2}{U(A)-U(x)}} . $$
A: If you look at a quartic potential you will see that it is "flatter" at the origin (higher powers progressively approximate a square well). The point is not to blindly apply the formula but to take the lowest order terms, which is usually quadratic.
Note you can still find the period of oscillation, as is shown here
