If you by "cone" mean exactly that $A$ should be isomorpic to $\mathrm{gr}_{\mathfrak m}A$ it seems that the following is counterexample: Let $G=\mathbb G_m$, $A=k[x,y,z]/(x^2+y^3+z^5)$ with $tx=t^{15}x$, $ty=t^{10}y$ and $tz=t^{6}z$ (exponents chosen more or less at random). Then the tangent cone at the origin (the fixed point) has affine algebra $k[x,y,z]/(x^2)$ and hence is not isomorphic to $A$. This is just raising your cusp example one dimension so that it becomes normal.
Addendum: Turning to the modified question let's try for a toric example: $G=\mathbb G_m^2$ so we should look at a monomial subring of $k[x,x^{-1},y,y^{-1}]$ with monoid of monomials saturated (equivalently being the set of integer vectors in a rational cone). Unless I am mistaken $k[xy,x^2y,xy^2,x^3y,xy^3]$ is such a ring (if I have missed some monomial in the saturation it won't matter for my argument). We then have that $x^2y$ and $xy^2$ are elements in $\mathfrak m\setminus\mathfrak m^2$ (this is clear irrespective if I have missed some elements in the saturation) but their product $x^2y\cdot xy^2=(xy)^3\in \mathfrak m^3$ which means that the associated graded is not a domain and hence not isomorphic to $A$. (I think the general condition for being a cone is that the generators of the monoid of monomials should lie on an affine hyperplane which is anyway how I arrived at this example.)