How far out from the Sun is visible light still sufficient to read a book? Recent pictures from the New Horizons spacecraft, shown below, seem to indicate that, at Pluto's distance, we are entering a twilight zone, with a distinct lack of colors, although that could be due to the planet terrain itself or the camera used to take the picture. I will call Pluto a planet, I grew up being told it was one, so it's a habit.
                
             
         
How far out from the Sun is visible light still sufficient to read a book? Could we expect colors at this distance?
Obviously there is no sharp cutoff point, so I use the criterion, "still sufficient to read a book" as a rough indication of the solar luminosity available at the furthest distance possible. 
I ask this question from pure curiosity, as I am completely amazed that the Sun can still provide so much light for New Horizon's camera at a distance of  4, 787, 131, 862 kilometres (give or take a bit). 
Regarding colour, Chris White's comment clarifies the matter:

It's starting to become more appreciated, but isn't fully widespread, that all space exploration and astronomy images are grayscale. Different color channels are combined in post-production, either by scientists or by press offices, but they are always taken separately. Indeed, all digital consumer cameras are grayscale imagers too, just they automatically take three filtered images and combine them for you, to make your life easier. 

NASA's Pluto time calculator from zibadawa timmy in  comments below.
 A: This is very rough and based on eyeballing without careful measurements:
I've got a four-watt nightlight.  I can read by it (not comfortably) at a distance of about a meter.  The sphere of radius 1 meter has a surface area of about 12 square meters, so it appears that 1/3 of a watt per square meter will (barely) suffice for reading.
The earth gets about 1400 watts per square meter from the sun.  This falls off, of course, like the square of the distance, which means Pluto (if I've done this right) should get about 1 watt per square meter, or about three times what I get when I'm reading uncomfortably 1 meter in front of my nightlight.  If you multiply the distance to Pluto by about $\sqrt{3}\approx 1.7$, you'll get to a place (about 8 billion km out) where you're down to what I get from my nightlight. 
A: One point, the difficulty of seeing colors in dim light is due to properties of the human vision system. Most cameras will not have the same effect and will be able to show vivid colors in even dim light (as long as the light is sufficient for imaging).
But as a good guess, with accommodation, you can read (to some extent) under a full moon. The sun (according to Wikipedia) is about $4\times10^5$ times brighter than the full moon here on earth.
By the inverse square relation, if we were at a distance $\sqrt{4\times10^5}$ further away from the sun than the earth, the intensity would be about that of our full moon. That works out to a distance of $\sim 630\ \text{au}$ or $9.4\times10^{10}\ \text{km}$. That's much farther than Pluto (which has a maximum distance from the sun of about $50\ \text{au}$. 
It appears that Phil Plait had a Bad Astronomy article on the sun from Pluto, where he mentions that on average it would be about 250 times brighter than the full moon on earth. I think that agrees with the above. Bad Astronomy Blog
A: The unit of illumination is the lux, lumens per square meter.


*

*What is the minimum lux required for reading?

*How many lux does the Sun provide at distance D?


What is the minimum lux required for reading?
You can plug all sorts of numbers into this depending on how good your eyes are, how big the print is, and how close you hold it to your face. I'm going to use civil twilight which is 3.4 lux. Others may like 1 foot-candle, 1 lumen at one foot, or about 10.764 lux. Whatever value you prefer, you can plug it into the formula below.
How many lux does the Sun provide at distance D?
To calculate the lux the Sun puts out we first have to calculate how much solar radiation (raw energy) it puts out at distance D? You take the luminosity of the Sun and distribute it over the surface of a sphere of radius D.
$$ \text{solar radiation in} \, \frac{Watts}{m^2} = \frac {3.846 \times 10^{26} W}{4 \pi D^2} $$
Not all of that is visible light. We need to convert that into lux, lumens per square meter. To do that you need to integrate the power output over the curve of visible light output using the Luminosity Function which, fortunately, someone has already done coming out with  a luminous efficacy of 93 lumens per W.
$$ \text{illumination in lux at} \, D = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi D^2} $$
At the Earth where $ D = 1.5 \times 10^{11} \, m $ we get $ 1.27 \times 10^{5} \, lux $. This is higher than direct sunlight at the Earth's surface because it does not account for atmosphere. This is good because New Horizons is in a vacuum.
How many lux are available to New Horizons?
Pluto is currently 32.6 AU from the Sun, or $ D = 4.89 \times 10^{12} \, m $. Plugging that in we get $ 1.19 \times 10^2 $ or $ 119 \, lux $. Plenty!
Or is it? That's how much light New Horizons is receiving from the Sun, but how much is bouncing off Pluto? Pluto has an albedo of about .6 so a little more than half the Sun's light is reflected back to New Horizons, or about 70 lux which is about the same as your average office hallway. Not great, but plenty for a long exposure.
This paper on LORRI agrees, "At Pluto encounter, 33 AU from the Sun, the illumination level is ~1/1000 that at Earth".
How far out from the Sun is visible light still sufficient to read a book?
We need to solve for D.
$$ I = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi D^2} $$
$$ I D^2 = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi} $$
$$ D^2 = \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi I} $$
$$ D = \sqrt{ \frac{93 \frac{lumens}{W} \times 3.846 \times 10^{26} W}{4 \pi I}} $$
Plug in $I = 3.4 \, lux$ and get $D = 2.8 \times 10^{13}$ or $186 AU$ which puts us past the Kuiper Belt and well into the Scattered Disc.
When using a different number for reading light note that lux changes with the square root of the distance. If you double the lux required you decrease the distance by 1.4. If we triple I to 10 lux (aka the foot-candle), the distance drops by 1.7 to about 108 AU. Still very far out.
Could we expect colors at this distance?
At Pluto you will see colors fine.  At 186 AU you will see colors about as well as you can see colors at civil twilight.
New Horizons has two instruments measuring visible light. LORRI is a long range panchromatic camera, meaning it acts like a regular digital camera and captures an approximation of what the human eye sees.
The other is the Ralph telescope. It is a multispectral visible and infrared imager meaning it takes multiple images at several different wavelengths. These will appear grey, the grey is a measurement of the light intensity at a specific wavelength. This is usually how spacecraft "see" color because scientists aren't interested in pretty pictures with multiple wavelengths smeared together, they want data about specific wavelengths. NASA public relations people mix the images together to approximate what matches what the human eyeball would see for press releases. They can't always get it quite right. Phil Plat discusses it in detail with the Mars landers.

To make these single-filter images into a color composite is not easy. If the red filter lets in less total light than the blue, you need to compensate for that when you add the images together. If the red filter is wider (lets in a wider range of reds) than the blue filter, you have to compensate for that, and so on.

Short version: pictures from LORRI are closer to what you would see than from Ralph.
A: Based on the information here which claims:


*

*Daylight is between $10^4$ to $2.5 \times 10^4\,\mathrm{lux}$, and

*1 candle at 1 foot is $10$ lux (I'll use this as the readability limit)


using the $1/r^2$ scaling yields that "daylight" will fall to $10\,\mathrm{lux}$ at somewhere between $30\,\mathrm{au}$ and $75\,\mathrm{au}$; Pluto is at around $40\,\mathrm{au}$, so its orbit is the correct order of magnitude for where the light intensity falls below "reading" levels.
By doing the analysis in terms of illuminance, this avoids some of the efficiency problems involved in doing the analysis based on watts. Note that lux is defined in terms of lumen, which accounts for human perception too.
