The 9 Billion Names of God

Ruby, 46

?A.upto(?M*9){|s|s[/(.)\1{3}|[N-Z]/]||puts(s)}

My original, similar solution was longer and wrong (it output base13 numbers, which isn't quite all of them due to leading zeroes), but I'll leave it here because it got votes anyway.

1.upto(13**9){|i|(s=i.to_s 13)[/(.)\1{3}/]||puts(s)}

C 140 177 235

Good old procedural style, no fancyness.
It counts (no write) 11,459,252,883 names in 8 minutes.
Next edit with the runtime and size of names file. Watch the sky...
Runtime 57 minutes, file size 126,051,781,713 (9 chars+crlf per row). Please tell me the monks' email address, so that I can send them the zipped file, for manual check...

Edit Golfed a little more, reworked the check for repeated letters.
Still not the shortest, but at least this one terminates and generates the required output.
Runtime 51 min, file size 113,637,155,697 (no leading blanks this time)

A side note: obviously the output file is very compressible, still I had to kill 7zip, after working 36 hours it was at 70%. Weird.

char n[]="@@@@@@@@@@";p=9,q,r;main(){while(p)if(++n[p]>77)n[p--]=65;else for(r=q=p=9;r&7;)(r+=r+(n[q]!=n[q-1])),n[--q]<65&&puts(n+q+1,r=0);}

Ungolfed

char n[]="@@@@@@@@@@";
p=9,q,r;
main()
{
    while (p)
    {
        if (++n[p] > 77)
        {
            n[p--] = 65; // when max reached, set to min and move pointer to left
        }
        else 
        {
            for (r=q=p=9; r & 7 ;) // r must start as any odd number
            {
                r += r+(n[q]!=n[q-1])); // a bitmap: 1 means a difference, 000 means 4 letters equal
                n[--q] < 65 && puts(n+q+1,r=0);
            }
        }
    }
}

Golfscript, 58 47 characters

"A"13
9?,{13base{65+}%n+}%{`{\4*/,}+78,1/%1-!},

Thanks to Peter Taylor, I am spared from the seppuku from not beating the Ruby solution! Run the code up to 10 yourself, and here is proof it skips the four-in-a-row numbers.