With a big enough bucket of no precise size, I could do this.
Let's start by calling the length of the cube's sides $\large\bf{s}$.
Tilt the cube so that three corners are the same height above the ground, and a fourth points straight down. (The other 4 corners mirror this; one will point straight up, the last 3 will be at a single height above the ground.)
The three equal height corners and the downward-pointing corner form an inverted triangular pyramid. Its base is an equilateral triangle whose sides are the diagonals of the three downward-facing cube sides. As diagonals of a cube face, the sides of the pyramid's base each have length $\bf{s\sqrt2}$. The (non-base) sides of the pyramid are the cube sides between the three equal height corners and the bottom-most corner; since they are just cube sides, their lengths are each $\bf{s}$.
Now partially fill the cube with water
until the water is level with the base of your inverted pyramid. How much water does this take?
The volume of a triangular pyramid is $\bf{V=\frac13 Bh}$, where $\bf{B}$ is the area of the pyramid's base and $\bf{h}$ is the pyramid's height.
The base of our pyramid is an equilateral triangle. We'll find many useful equations at Wikipedia: for example, we know that the area is found by using $\bf{A=\frac14a^2\sqrt3}$ where $\bf{a}$ is the side length. In our case, $\bf{a=s\sqrt2}$, so $\bf{A=\frac14(s\sqrt2)^2\sqrt3 = \frac14(2s^2)\sqrt3 = \frac{s^2\sqrt3}2}$, giving
$$\bf{B=\frac{s^2\sqrt3}2}$$
The height of our pyramid extends from the midpoint of its base perpendicularly to the apex corner. The midpoint of the base triangle sits thusly, where $h_a$, $h_b$, and $h_c$ intersect:
For ease of reference, the base midpoint (where $h_a$, $h_b$, and $h_c$ meet) we'll call $\bf{M}$, and the midpoint of side $AB$ we'll call $\bf{P}$.
The pyramid's height $\bf{h}$ will be the third side of a triangle whose other two sides are an edge of the pyramid (e.g. between $B$ on the diagram and the pyramid's apex), and the line between that same edge and the midpoint of the base (e.g. between $B$ and $\bf{M}$). This is a right triangle, as the pyramid's height is perpendicular to its base.
Wikipedia tells us $\beta$ is 60°; from this, we see a 30°/60°/90° triangle on the diagram: $\Delta{B\bf{PM}}$, with $\angle{\bf{M}}B\bf{P}=30°$ and $\angle{\bf{PM}}B=60°$. Wikipedia also tells us the height from the center of each side to $\bf{M}$ is $\bf{\frac{h}3}$ where $\bf{h=\frac{\sqrt3}2a}$. Here, $\bf{a}$ is (still) $\bf{s\sqrt2}$, so we find $\bf{\overline{PM}=\frac{h}3=(\frac13)(\frac{\sqrt3}2a)=(\frac13)(\frac{\sqrt3}2s\sqrt2)=\frac{s\sqrt6}6}$, so by the ratio for 30°/60°/90° triangles, we know $\overline{B\bf{M}}\bf{=\frac{s\sqrt6}3}$.
We can now find $\bf{h}$ as we know the other two sides of the right triangle it is in.
- Side 1 is $\bf{h}$.
- Side 2 is $\bf{=\frac{s\sqrt6}3}$.
- Hypotenuse is $\bf{s}$, as it is a pyramid (non-base) edge and thus a cube side.
Pythagorean's now gives us
$$\bf{h=\sqrt{s^2-(\frac{s\sqrt6}3)^2}=\sqrt{s^2-\frac69s^2}=\sqrt{\frac{s^2}3}=\frac{s}{\sqrt3}}$$
On the home stretch now ...
$\bf{V=\frac13 Bh = (\frac13)(\frac{s^2\sqrt3}2)(\frac{s}{\sqrt3})=\frac13(\frac{s^3}2)=\frac12(\frac{s^3}3)}$
Now, we wanted to fill exactly $\bf{\frac13}$ the cube's volume with water. And we hopefully know that the cube's volume is simply $\bf{s^3}$. So we want an amount of water equal to $\bf{\frac{s^3}3}$.
We just found our pyramid has exactly half the desired amount!
- We just partially filled the cube with a volume of water $\bf{\frac12(\frac{s^3}3)}$.
- Pour that water into our arbitrarily sized but big enough bucket.
- Partially fill the cube in exactly the same way as before.
- Adjust the cube so the hole is on top.
- Pour the contents of the bucket carefully into the hole.
You now have the cube filled one third of its own volume with water.
