Can a molecule be neither gerade nor ungerade?
g/u is not a property of a molecule; I assume you meant molecular orbital.
All functions can be written as a linear combination of even + odd functions: there is a short explanation on Wikipedia. Basically, you have a function $f(x)$; now define 

$$\begin{align}
f_\mathrm e(x) &amp;= \frac{1}{2}[f(x) + f(-x)] \\
f_\mathrm o(x) &amp;= \frac{1}{2}[f(x) - f(-x)] \\
\end{align}$$

It is clear from the definition of parity that $f_\mathrm e$ is even (since $f_\mathrm e(x) = f_\mathrm e(-x)$) and likewise that $f_\mathrm o$ is odd. Now since $f(x) = f_\mathrm e(x) + f_\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.

In three dimensions where you have $\psi(x,y,z)$ you simply need to define

$$\begin{align}
\psi_\mathrm g(x) &amp;= \frac{1}{2}[\psi(x,y,z) + \psi(-x,-y,-z)] \\
\psi_\mathrm u(x) &amp;= \frac{1}{2}[\psi(x,y,z) - \psi(-x,-y,-z)] \\
\end{align}$$

and analogously to above one can see that every function $\psi(x,y,z)$ can be expressed as the sum of a gerade component $\psi_\mathrm g$ and an ungerade component $\psi_\mathrm u$. These components don't have any physical meaning (unless $\psi$ itself is either g or u, in which case one component is simply $\psi$ and the other is zero), but that's somewhat beside the point.

If a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.

So the answer is yes, and yes. There is no "or".