The answer is order 2.  Here is why:

First of all, the equation of a parabola with any horizontal 
axis of symmetry y = k is

(y-k)² = 4p(x-h)

That has three arbitrary constants h, k and p.

Therefore a differential equation of all parabolas 
having ANY horizontal axis of symmetry (not necessarily the
x-axis, would be of order 3, the number of arbitrary constants.

If we restrict the family of parabolas to those with axis of 
symmetry y = 0, (where k = 0), the equations would be

y² = 4p(x-h)

y² = 4px - 4ph

or 

y² = c1x + c2, where c1 = 4p and c2 = 4ph

That has 2 arbitrary constants, so the order would be 2.

That's the answer you were looking for.  But I'll continue:

To find that differential equation, we differentiate
twice to get rid of the arbitrary constants:

y² = c1x + c2

2y*y' = c1

 y*y' = c   where c = c1/2 

y*y" + y'*y' = 0

y*y" + (y')² = 0

That would be the 2nd order differential equation whose solution 
is all parabolas with their axis of symmetry being the x-axis.

Edwin