I don't know of any research to find out if skin sunburns faster when wet, though someone did a comparable experiment to find out if plants can be burnt by sunlight focussed through drops of water after the plants have been watered.
You need to be clear what is being measured here. The total amount of sunlight hitting you, and a plant, is unaffected by whether you're wet or not. The question is whether water droplets can focus the sunlight onto intense patches causing small local burns.
The answer is that under most circumstances water droplets do not cause burning because unless the contact angle is very high they do not focus the sunlight onto the skin. Burning (of the plants) could happen if the droplets were held above the leaf surface by hairs, or when the water droplets were replaced by glass spheres (with an effective contact angle of 180º).
My observation of water droplets on my own skin is that the contact angles are less than 90º, so from the plant experiments these droplets would not cause local burning. The answer to your question is (probably) that wet skin does not burn faster. I would agree with Will that the cooling effect of water on the skin may make you unaware that you're being burnt, and this may lead to the common belief that wet skin accelerates burning.
I am wondering whether an object which has been wet with hot water always dries up more easily and more quickly than one wet with cold water.
Yes, it will. The more important question is, "How much better will it dry?".
Some simple intuition may be garnered from dishwashers. When a thick ceramic plate or bowl goes through a dishwasher, it winds up hot and covered in water by the end of its journey. When you open the dishwasher and wait for 15 minutes, you will find the thick plate or bowl will be almost completely dried. For a thin plastic dish, something entirely different happens: it still remains mostly wet by the time it has cooled down.
Why? The thick ceramic bowl has a greater thermal mass per surface area than the thin plastic plate, where thermal mass is defined to be $m C_p$ with $C_p$ being the constant pressure heat capacity. The thick bowl holds enough heat to evaporate its coat of water and dry in the presence of air; the thin plate simply doesn't hold enough heat to drive off its water.
Ignoring heat loss due to radiative effects, the ability of an object to hold enough heat to drive off its water is entirely determined by the ratio of its thermal mass to its water content. To be specific, the fraction of water which will be lost is approximately given by
$$x=\frac{m_oC_p\Delta T}{m_w\Delta H}$$
where $m_o$ is the object mass, $m_w$ is the water mass adsorbed, $\Delta T$ is the difference between the object initial temperature and ambient temperature, $C_p$ is the heat capacity of the object (not the water), and $\Delta H$ is the latent heat of vaporization of water in units of energy per mass per Kelvin, which for water happens to be 2260 Joules per gram per Kelvin.
For an object with a typical heat capacity of 1 Joule/gram/Kelvin with 5% water content heated initially to 373K, a drop to room temperature will be achieved with about 60% water loss. So your object had better be less than 3% water if you want it to dry effectively just from its thermal heat.
Best Answer
Evaporation of sweat from the skin surface has a cooling effect due to the latent heat of evaporation of water (from wikipedia). Basically, the heat is transfered from your skin to the water, the water then evaporates, taking with it a bit of the energy stored in it.
Further references here