They are sometimes also called double planets and they're more widespread in fiction than in observations. I don't think that there is any new instability that would appear for the system of double planets orbiting a star and that wouldn't be present for other, more asymmetric pairs of planets. Obviously, the tidal forces would be really large if the planets were close enough to each other. But because the tidal forces go like $1/r^3$, it's enough to choose the distance that is 5 times larger than the Earth-Moon distance and the tidal forces from the other Earth would be weakened 125 times and would already be as weak as they are actually from the Moon now (with a lower frequency).
One must realize that the systems with two or several planets are rather rare and the condition that the mass of the leading two planets is comparable is even more constraining.
Imagine that each of the two planets has a mass that is uniformly distributed between 1/20 of Earth's mass (like Mercury) and 300 Earth masses (like Jupiter) on the log scale. The interval goes from the minimum to the maximum that is 6,000 times heavier. That's more than 12 doublings, $2^{12}=4,096$. So if you pick the first planet to be at a random place on that interval of masses (uniformly at the log scale), the probability that the second planet's mass (which is independent) differs by less than the factor of $\sqrt{2}$ from the first mass is about $1/12$.
Only $1/12$ of systems that look like a pair of planets will be this symmetric. And the number of pairs of planets - even asymmetric ones – is rather low, indeed. The reason is really that during the violent eras when Solar-like systems were created, rocks had large enough velocities and they flew in pretty random directions so they were unbound at the end. It's just unlikely to find two large rocks in the same small volume of space: compare this statement with some high-temperature, high-entropy configurations of molecules in statistical physics.
It's also rather unlikely that a collision with another object creates two objects that will orbit one another. After all, the two-body orbits are periodic so if the two parts were in contact during the collision, they will collide again after one period (or earlier). Equivalently, the eccentricity of the orbit is likely to be too extreme which will lead to a fast reunification of the two new planets. Moreover, even if something would place the two newly created planets from a "divorce" on a near-circular orbit, perhaps a collision with a second external object (good luck), it's very unlikely that such an orbit will have the right radius, like the 1 million km I was suggesting in the case of the hypothetical double Earth above. If the two objects are too close, the tidal forces will be huge and (at least for some signs of the internal angular momentum) they will gradually make the planets collapse into one object again. And if the energy with which the planets are ejected from one another is too high, no bound state will be created at all. So the initial kinetic energy of the newborn 2 planets would have to be almost exactly tuned to their gravitational potential energy (without the minus sign) and that's generally unlikely, too.
According to E. V. Petjeva (2011), the measured rate of change of the Earth-Sun distance (astronomical unit) is (1.2 +/- 3.2) cm/yr, with the uncertainty value representing 3 standard deviations. In other words, any change is within the uncertainty of the measurement. She specifically addresses the Krasinsky and Brumberg value. E. M. Standish has also addressed this issue.
The measurements are by radar echos off other planets and radio signals from space craft. See the below references for details.
(Note that "astronomical unit" is not technically the same as Earth-Sun distance, see Standish reference for details, Krasinsky and the others are all measuring the "astronomical unit"; also "astronomical unit" was redefined to be a constant in 2012, see Nature reference for details)
http://syrte.obspm.fr/jsr/journees2011/pdf/pitjeva.pdf
http://adsabs.harvard.edu/abs/2005IAUCo.196..163S
http://www.nature.com/news/the-astronomical-unit-gets-fixed-1.11416
Best Answer
This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens.
Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$. You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$. Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$. Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$. The time it would take for this to make a significant dint the the Earth's orbital velocity ($30\ \mathrm{km/s}$) is of the order of $10^{15}\ \mathrm{yr}$. I think we're safe.
For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $ R r^{3/2} $, which for Mercury is about 0.1. So the end result is not much different for Mercury.
This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in:
Note that the number of digits displayed in the final column is ludicrous. :)