Overloading DC transmission wire
Your reasoning is partly correct, and partly makes no sense at all.

AWG 11 for 12 A is very conservative, considering 10 gauge is approved for 30 A in house wiring.

Look up the resistance per unit length of wire for the different wire sizes.  The power dissipation over that same length will be that resistance times the square of the current.  For example, if some wire has 100 m&Omega; resistance per foot, then with 12 A thru it the dissipation will be (12 A)2(100 m&Omega;) = 14.4 W, or 1.2 W/inch.  That's going to get quite toasty and probably requires high temperature insulation and making sure the wire is kept away from flammable things, like the dry wooden supports of a house.

The part of your statement that makes no sense is saying that you will get "4 amp loss of electricity".  That's wrong in so many ways.  First, the current thru the wire is all the same.  The one thing you don't "loose" is current.  Second, "electricity" isn't a quantity or something specific.  It's like saying today you'll have more weather, as opposed to snow, rain, etc.

What you will lose is a voltage along the end of the wire.  The amount of voltage dropped by the wire is its resistance times the current thru it.

If you really need to use thin wire and still want to transfer substantial power, use a higher voltage.  Send 100 V or something, then down-convert at the receiving end.  The extra switching power supply may well be cheaper than a thicker cable.

The reason higher voltage helps is because less current is required to carry the same power.  You say you want to transfer 300 W.  That only takes 3 A at 100 V, but 15 A at 20 V, for example.  Since the losses and requirement for wire size comes from the current, not the voltage, using high voltage and low current saves on wire and losses.  Of course if you go too high in voltage you have to consider safety and possibly extra insulation.  Physics can be annoying like that.

Added about 20 gauge

OK, I just looked up 20 gauge wire.  It has a resistance of 33.3 m&Omega;/m, which is 10.2 m&Omega;/foot.  At 12 A it will dissipate 1.5 W/foot, or 122 mW/inch.  That will get noticeably warm, but should be fine for most purposes.

Now let's look at the losses.  1000 feet distance means 2000 feet of wire, since the current has to flow out and back.  (2000 ft)(10.2 m&Omega;/ft) = 20.3 &Omega;.  At 12 A that will drop 244 V.  That's how much you have to apply to the two ends just to get 12 A thru the wire with a short at the other end.  If you want 12 V available at the other end, then you have to apply 256 V at the sending end.  It should be clear that this is horribly inefficient.

Use a thicker wire, a higher voltage, or both.