Your technology supposition is a huge one. Real fusion explosives must use small conventional and fission explosions to create the confinement and temperature needed to get a rapid fusion explosion. But, assuming that -- You can calculate what is known as the Q of each reaction:
$$Q=m(reactants)-m(products)$$ so for your fusion reaction in the strongest (only?) channel we have$$Q_{fusion}=m(^3H)+m(^2H)-m(^4He)-m(^1n)$$ where the $m$'s are nuclear masses. Because all the missing electron masses will cancel, you can use atomic masses which are in tables here and here. A typical fission reaction would be
$$Q_{fission}=m(^{235}U)+m(^1n)-m(^{137}Cs)-m(^{96}Rb)-3m(^1n)$$
If you calculate the Qs for these reactions and compute the Q/m(reactants) for each you will see that the fusion event releases over 4$\times$ the energy per reactant mass.
In this scenario, the fusion event produces a helium nucleus and a neutron with no direct radioactive by-products. The fission event produces 2-3 neutrons, depending on the exact products (a bi-modal distribution and not a single channel), but your numbers will show that for equal energy output, the fusion event produces about 3$\times$ the neutrons.
The fission explosion would produce a wide variety of radioactive by-products along with a tremendous spectrum of x-rays and gamma rays. The neutron production in the fusion device could produce radioactive products via neutron activation reactions with the bomb casing and the atmosphere. The actual energy release may produce some high energy gamma rays due to excitation of atmospheric oxygen and nitrogen.
Another thing to keep in mind is that a very small fraction of the reactant ingredients in these devices actually react, so there will be leftover tritium in the fusion device and leftover uranium (or plutonium) in the fission device.
This is called a 'skirt' or 'bell' and it is indeed a condensation effect: humid air is entrained by the rising column, and water then condenses out as the pressure falls. These droplets, if they get big enough fast enough, then fall with respect to the rising air, resulting in these skirts.
Condensation phenomena are fairly common with nuclear (and other large) explosions.
This is covered in the Wikipedia article.
Best Answer
Mushroom clouds are formed in explosions (not necessarily nuclear - see picture) as a result of the Rayleigh-Taylor instability. For given density contrast $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}$ between the hot cloud and the cold atmosphere, the timescale $t_{RT}$ for this instability scales with the length scale $L_{RT}$ according to:
$$t_{RT} \approx \sqrt{\frac{L_{RT}}{g}\frac{\rho_{cold}+\rho_{hot}}{\rho_{cold}-\rho_{hot}}}$$
With $g$ the gravitational acceleration. For $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}\approx 0.1$ and $L_{RT} = 1 km$ we find $t_{RT} \approx 30 s$. For a cup sized ($L_{RT} = 0.05 m$) explosion with the same density contrast we find $t_{RT} \approx 0.2 s$.
The 'mushroom' has disappeared before it really takes shape. Bottom line is that you need fairly large explosions to observe a mushroom.