It all depends on your interpretation (setup).
If your goal is to compute the $2$-dimensional area under the curve $y = x^2$ then $y$ must also have the unit of metre ($m$), which gives an area in terms of square-metre ($m^2$), as you would expect.
If, however, you are aiming for a "natural" 3-dimensional geometric interpretation, then you can think of it like so: The $y$-value gives the enclosed area of a square of side $x$. So the integral you computed actually gives you the enclosed volume of a square pyramid. If you take cross-sections across such a shape, you will get squares of different sides ($x$) and different enclosed areas $(y = x^2)$. Naturally, the calculated volume will have units of cubic-metre $(m^3)$. If you start the $x$ value off at something other than zero, you will be computing the volume of a square-pyramidal frustum (truncated square pyramid).