Answer: Yes.

You can re-arrange the equation as:

3(a+b+c+d)(a+b-c-d) = (a-b+c-d)(-a+b+c-d)

So if a+b+c+d is prime then either a-b+c-d or -a+b+c-d is a multiple of a+b+c+ d. 

But since they are both smaller than a+b+c+d in magnitude, then one of them must be zero. Since the right-hand side of the equation is zero, so is the left hand side. That means a+b-c-d=0, so a+b=c+d, so a+b+c+d=2(a+b), which is an even number bigger than 2,
proving the statement.

How the heck did I ever manage to rearrange

a²+ab+b²-c²-cd-d²=0

to get

3(a+b+c+d)(a+b-c-d) = (a-b+c-d)(-a+b+c-d)  ?

The trick was to write:

a² + ab +

b² = (3(a+b)² + (a-b)²)/4

and find a similar expression for

c²+cd+d², equate these appropriately, regroup then use difference of two squares.

Note, this equation is in effect a diagonalization of the quadratic form a²+ab+b². That is if we define new variables X=a+b, Y=a-b then the quadratic form becomes 3X²/4 + Y²/4 which is diagonal because there are no cross terms -- that is, the coefficient of XY is zero.