The heat transfer from 'hot' air or water to 'cold' sausages is roughly determined by Newton's law of cooling/heating:
$$\boxed{\frac{\text{d}Q}{\text{d}t}=hA[T_{\infty}-T(t)]}\tag{1}$$
where, for heating:
$\frac{\text{d}Q}{\text{d}t}$ is the rate of heat transfer into of the body, which determines how quickly the body's temperature rises,
$h$ is the heat transfer coefficient (assumed independent of T and averaged over the surface),
$A$ is the heat transfer surface area,
$T(t)$ is the temperature of the object's surface, as a function of time $t$,
$T_{\infty}$ is the temperature of the environment; i.e. the temperature suitably far from the surface.
For now we'll assume that $T_{\infty}=\text{constant}$, throughout the heating process.
In that case, acc. $(1)$, the rate of heating depends strongly on $h$, the convection coefficient. It's well known that broadly speaking $h$ is much larger for liquids than for gases, all other things being equal. This table of $h$ values bears that out. So based on this it's reasonable to assume that defrosting sausages in water will be quicker than in air.
There is however a caveat. With water heating not only will the temperature of the sausages increase but also will the temperature of the water decrease. The latter, according to $(1)$, will decrease the heating rate. The trick to avoid this is to simply uses a large ratio of water to sausages or simply run the sausages under the cold faucet, continuously.