The transfer function is not \$H(\omega)\$, it is \$H(j\omega)\$ (note the \$j\$, which makes it complex):
$$H(j\omega) = \frac{1}{1+j\omega RC}$$
This is important because the transfer function captures the phase in addition to the amplitude.
The amplitude is
$$|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}$$
by the definition of complex magnitude. This is the gain you measure from input to output and it may be all you care about, especially for a first order system. However, the phase $$\angle H(j\omega) = -\arctan(\omega RC)$$ can be quite important (e.g. for ensuring stability), especially for higher order transfer functions.
The main, direct use of the transfer function is to capture both gain and phase in one expression, but can also be used for time-domain analysis, as the transfer function is the Laplace transform of the impulse response. It is also useful for characterizing a multiple stage system, since the transfer function \$H(j\omega)\$ of a system consisting of stage \$H_1(j\omega)\$ followed by \$H_2(j\omega)\$ is simply \$H(j\omega) = H_1(j\omega)H_2(j\omega)\$ whereas in the time-domain you need to convolve the impulse responses \$h_1(t)\$ and \$h_2(t)\$ to find the overall system impulse response of \$h(t)\$.
If \$\omega_0\$ equals \$\dfrac{1}{RC\sqrt2}\$ then what's the problem?
This would make Q = \$\dfrac{1}{\sqrt2}\$.
It all sounds very reasonable to me. DC transfer function will be 0.5 as indicated by @LvW.
Best Answer
No - the transfer function of a filter does not depend on the selected topology. However, this applies to IDEAL conditions only. That means: Ideal properties of the active elements as well as for all passive parts (no tolerances).
If these requirements of ideality are not "sufficiently" fulfilled ("sufficiently" means: with acceptable error) different topologies will exhibit different deviations from the desired transfer function. This is the most important criterion for selecting one of the various topologies for a certain application. As an example, the well-known Sallen-Key topology is less sensitive to the opamps finite open-loop gain values (frequency dependent gain) but pretty sensitive to parts tolerances. In this respect, the so called multi-feedback topology (MFB) has opposite properties.
To select a suitable topology for a specific application is a rather challenging task for an engineer bacause he has to take into account technical as well as economical aspects.
In short: Under ideal conditions all topologies give the same filter behaviour - however, because nothing is ideal, it is important to select the "proper" method for realizing the filter. In particular, this applies to higher-order filters because, in these cases, we have to select between two principles: Series connection of active second-order stages or active realzation of passive RLC structures (Keywords: Leap frog, active L-simulation, FDNR technique).
EDIT/UPDATE: For better explaining the influence of non-ideal opamp data on filter performance I like to mention the output resistance of the opamp. Normally, this parameter is always neglected during calaculation (set to zero). However, the Sallen-Key lowpass structure has one severe drawback: For rising frequencies there is a "direct" path between input and opamp output via the feedback capacitor. As a result the damping properties far above the cut-off frequency are heavily disturbed. This is due to a voltage drop across the finite output resistance of the opamp caused by the current that directly goes to the output node.