Scenario 1: The scenario where Jupiter just drops into the Sun from its current position would certainly have short-term effects. But short-term here means compared with the lifetime of the Sun, not hundreds of years. NB I have not yet (fully) considered an alternate scenario where Jupiter is drawn closer and loses its energy and angular momentum gradually as it gets nearer and nearer to the Sun.
The kinetic energy of Jupiter at the Sun's surface would be of order $GM_{\odot}M_{\rm Jup}/R_{\odot} \sim 10^{39}$ Joules.
The solar luminosity is $3.83 \times 10^{26}$ J/s.
The addition of this much energy (if it is allowed to thermalise) would potentially affect the luminosity of the Sun for timescales of tens of thousands of years.
However, I suspect what is more likely is that the kinetic energy would be also be used to do work and lift the convective envelope of the Sun. In other words, the Sun would both increase in luminosity and in radius. If the effects were just limited to the convective envelope (it is unclear whether the plummeting planet could survive falling to even greater depths), then this has a mass of around $0.02 M_{\odot}$ and so could be "lifted" by $\sim 10^{39} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.1 R_{\odot}$
So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^{5}$ years.
On longer timescales the Sun would settle back down to the main sequence, with a radius and luminosity only slightly bigger than it was before.
This all assumes that Jupiter can remain intact as it falls. It certainly wouldn't "evaporate" in this direct infall scenario, but would it get tidally shredded before it can disappear below the surface? The Roche limit is of order $R_{\odot} (\rho_{\odot}/\rho_{\rm Jup})^{1/3}$. But the average densities of the Sun and Jupiter are almost identical. So it seems likely that Jupiter would be starting to be tidally ripped apart, but as it is travelling towards the Sun at a few hundred km/s at this point, tidal breakup could not be achieved before it had disappeared below the surface.
So my conclusion is that dropping Jupiter into the Sun in this scenario would be like dropping a depth charge, with a lag of order $10^{5}$ years before the full effects became apparent.
Scenario 2: Jupiter arrives at Roche limit (just above the solar surface) having mysteriously lost a large amount of angular momentum.
In this case what will happen is Jupiter will be shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of $2 R_{\odot}$, the orbital period will be about 8 hours, the orbital speed about 300 km/s and the orbital angular momentum about $2\times 10^{42}$ kg m$^2$ s$^{-1}$. Assuming total destruction, much of the material will form an accretion disc around the Sun, since it must lose some of its angular momentum before it can be accreted.
How much of the Sun's light is blocked is uncertain. It mainly depends on how the material is distributed in the disk, especially the disk scale height. This in turn depends on the balance of the heating and cooling mechanisms and hence the temperature of the disk.
Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and $2R_{\odot}$ and that it gets close to the solar photospheric temperature at $\sim 5000$K. In which case the disk area is $3 \pi R_{\odot}^2$, with an "areal density" of $\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$.
In hydrostatic equilibrium, the scale height will be $\sim kT/g m_H$, where $g$ is the gravitational field and $m_H$ the mass of a hydrogen atom. The gravity (of a plane) will be $g \sim 4\pi G \sigma$. Putting in $T \sim 5000$K, we get a scale height of $\sim 0.06 R_{\odot}$.
Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $\sim 20$%, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.06 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 10$ kg/m$^3$. Looking though this corresponds to a column density of $\sim 10^{10}$ kg/m$^{-2}$.
For comparison, the solar photospheric density is of order $10^{-12}$ kg/m$^{3}$ and is only the upper 1000 km of the Sun. Given that the definition of the photosphere is where the material becomes optically thick, we can conclude that a tidally shredded Jupiter is optically thick to radiation and indeed the sunlight falling on the Earth would be very significantly reduced.
How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure.