That such a decomposition doesn't exist can be shown by looking at the interior solid angles at the vertices.
The cube has eight interior solid angles of $\pi/2$ each.
Each of the six tetrahedra has one interior solid angle of $\pi/2$ and three that can be computed as shown here:
$$\Omega=\phi_{ab}+\phi_{bc}+\phi_{ac}-\pi=\pi/2 + \arccos\frac{1}{\sqrt{3}}+ \arccos\frac{1}{\sqrt{3}}-\pi=2\arccos\frac{1}{\sqrt{3}}-\pi/2\;.$$
Thus, the sum of the interior solid angles of the cube is $4\pi$, and the sum of the interior solid angles of the tetrahedra is
$$6\left(\pi/2+3\left(2\arccos\frac{1}{\sqrt{3}}-\pi/2\right)\right)=6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)\;.$$
This is not equal to $4\pi$, since $\arccos\frac{1}{\sqrt{3}}$ is not a rational multiple of $\pi$. (Here's an elementary proof of that fact.) In fact it is almost $5\pi$.
[Edit: I just realized in the shower that all the following analysis of potential inner vertices is actually not necessary. No matter what happens at the inner vertices, we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle $\pi/2$ are not enough to do it, so we have to use at least one of the others, but then we can't make the sum come out to $4\pi$, a rational multiple of $\pi$.]
This establishes that you can't fill the cube with the six tetrahedra by letting all their vertices coincide with the cube's vertices. It seems geometrically obvious that you can't do it with any of the vertices inside the cube, either, but this, too, can be proved rigorously using the solid angles. If there were a vertex inside the cube, the entire solid angle of $4\pi$ around that inner vertex would have to be filled. The candidates for filling it are a face of a tetrahedron, which subtends a solid angle of $2\pi$, an edge of a tetrahedron, which subtends a solid angle of twice the dihedral angle of the intersecting planes, i.e. either $\pi$ or $2\arccos\frac{1}{\sqrt{3}}$, or an interior solid angle of a vertex of a tetrahedron. Thus, adding up all the solid angles at the eight outer vertices and at $v$ inner vertices, the following equation would have to have solutions with non-negative integer values of $j$, $k$, $l$, $m$, $n$ and $v$:
$$6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)+j(2\pi)+k(\pi)+l\left(2\arccos\frac{1}{\sqrt{3}}\right)+m(\pi/2)+n\left(\arccos\frac{1}{\sqrt{3}}-\pi/2\right)$$
$$=4\pi+v(4\pi)\;.$$
This is impossible, since the coefficient in front of $\arccos\frac{1}{\sqrt{3}}$ is non-zero, and thus the equation would imply that this is a rational multiple of $\pi$, which it isn't (see above).
P.S.: I just realized I forgot to mention an essential part of the proof:
$$\cos 2\phi = 2 \cos^2 \phi - 1\;,$$
$$\arccos x = \frac{1}{2}\arccos(2x^2-1)\;,$$
$$\arccos \frac{1}{\sqrt{3}}=\frac{1}{2}\arccos(-\frac{1}{3})\;,$$
so $\arccos \frac{1}{\sqrt{3}}$ is rational iff $\arccos(-\frac{1}{3})$ is rational; then we can apply the theorem I linked to above.
A cube actually has eight corners, not four.
It has twelve edges, not eight.
The center of the sphere can only travel within a $3\times3\times3$ cube in the center of the $5\times5\times5$ cube. Whenever the center of the sphere reaches the surface of the $3\times3\times3$ cube, some part of the sphere touches the surface of the $5\times5\times5$ cube in such a way to prevent the center from moving out of the $3\times3\times3$ cube.
If you divide the $5\times5\times5$ cube into $1\times1\times1$ little cubes, there are eight little cubes, one in each corner, in each of which the sphere can only reach the parts it occupies when it is at one corner of the little cube
(which is also one corner of the $3\times3\times3$ cube).
At this point, just one eighth of the sphere is in the little cube;
the other $\frac78$ of the sphere is in the other seven little cubes around the center of the sphere.
Since the volume of the sphere is $\frac43 \pi r^3 = \frac43 \pi \times 1^3,$ one-eighth of the volume is $\frac18 \times \frac43 \pi \times 1^3.$
But this is the part of the little cube that is reachable by the sphere;
the part that is unreachable has volume equal to the little cube's volume minus the part occupied by the sphere. This is
$ 1 - \frac18 \times \frac43 \pi \times 1^3. $
There are eight little cubes in corners of the large cube that have this much unoccupied space, so their total unoccupied space is
$$ 8 \left(1 - \frac18 \times \frac43 \pi \times 1^3\right) . \tag1$$
Along each edge of the $5\times5\times5$ cube there are three other little cubes (not including the ones in the corner of the large cube, which we've already counted).
In each of these three little cubes, the sphere is able to reach all the points it can by sliding its center along one edge of the little cube (which is also part of an edge of the $3\times3\times3$ cube). If you do that, the sphere sweeps through a volume inside the cube that is one quarter of a cylinder of radius $1$ and height $1.$
The volume of a cylinder of radius $1$ and height $1$ is
$\pi r^2 h = \pi \times 1^2 \times 1.$
One quarter of the cylinder is
$\frac14 \pi \times 1^2 \times 1.$
The unreachable volume of the little cube is the volume of the little cube, $1^3,$ minus the volume that can be reached by the sphere, so the unreachable volume is
$1^3 - \frac14 \pi \times 1^2 \times 1.$
There are three little cubes along one edge of the large cube, so in these three little cubes the total unreachable volume is
$\left[1^3 - \frac14\pi \times 1^2 \times 1\right] \times 3.$
And this is repeated along each of the twelve edges of the large cube, so the total unreachable area in all those little cubes is
$$ 12\left(\left[1^3 - \frac14\pi \times 1^2 \times 1\right] \times 3\right). \tag2$$
There are no other unreachable parts of any little cubes, so to sum up all the unreachable volume we simply add the parts found on line $(1)$ and on line $(2)$.
In the one-line solution you found, for some reason $\pi \times 1^2 \times 1$ is simplified to $\pi \times 1^2$ but $1^3$ is not simplified to $1$ and other multiplications by $1$ are not simplified. But the underlying structure of the sum is clear enough.
Best Answer
I have no idea about any upper bound. However, for $d = 2$, $a$ is bounded from below by $\frac14$.
There is a result by Hadwiger${}^{\color{blue}{[1]}}$:
Let $K_1$ be your $F$ and $L$ be its perimeter. Since $[0,1]^2$ is convex and $F \subset [0,1]^2$, we have $V \le 1$ and $L \le 4$.
Let $K_2$ be a circle of radius $r \le \frac{V}{4}$. Since
$$2\pi(A_1 + A_2) = 2\pi(V+\pi r^2) \ge 2\pi V \ge 8\pi r \ge L(2\pi r) = L_1L_2$$ Condition $(*1)$ is satisfied. Since $A_2 = \pi r^2 \le \frac{\pi}{16}V^2 < V = A_1$, it is impossible to make $K_2$ contains $K_2$. Hadwiger's result tell us one can translate/rotate $K_2$ to make $K_1 = F$ contains $K_2$.
Since this is true for all $F$, we can conclude $a$ is bounded from below by $\frac14$.
Update
It turns out there is a more elementary derivation of a lower bound that can be generalized to higher dimensions. For simplicity of argument, we will assume $F$ is a closed polytope.
For $k \in \mathbb{Z}_{+}$ and geometric shapes $S \subset \mathbb{R}^d$, let $\mu_k(S)$ be the $k$-dim measure of $S$ whenever that make senses.
Let $\mathcal{X}$ be the collection of convex polytopes which are closed, bounded and has non-empty interior. It is known that any $X \in \mathcal{X}$ is a finite intersection of closed half-spaces. More precisely, let $\mathcal{F}$ be the set of facets of $X$. For each facet $f$, let $p_f$ be a point on $f$ and $n_f$ be the outward pointing unit normal vector. For any $s$, define $$H_f(s) = \{ x \in \mathbb{R}^d : n_f \cdot (x - p_f) \le s \} \quad\text{ and }\quad X(s) = \bigcap_{f \in \mathcal{F}} H_f(s)$$
For sufficiently small $s$, $X(s) \in \mathcal{X}$. In particular $X = X(0)$.
Let $\begin{cases} V(s) = \mu_d(X(s))\\ A(s) = \mu_{d-1}(\partial X(s)) \end{cases}$ be the hyper-volume/area of $X(s)$.
In particular, $V = V(0)$ and $A = A(0)$ are the hyper-volume/area of $X$.
Since $X$ is bounded, one can show
Furthermore, $V(s)$ is differentiable on $(-R,0)$ with $\frac{d}{ds}V(s) = A(s)$
Notice for $s \le 0$, $X(s) \subset X(0) = X$ and both of them are convex, we also have $A(s) \le A$ ${}^{\color{blue}{[2]}}$.
By continuity, we have $V(-R) = 0$. Combine these, we find
$$R \ge \frac1{A}\int_{-R}^0 A(s) ds = \frac1{A}\int_{-R}^0 \frac{d}{ds}V(s)ds = \frac{V(0) - V(-R)}{A} = \frac{V}{A} $$
Back to our original problem and let $X$ be your $F$.
Since $F \subset [0,1]^d$ and both $F$ and $[0,1]^d$ are convex, we have${}^{\color{blue}{[2]}}$
$$A = \mu_{d-1}(\partial F) \le \mu_{d-1}(\partial [0,1]^d) = 2d$$
For $r = \frac{V}{2d}$, we have $$r \le \frac{V}{A} \le R\quad\implies\quad X(-R) \subset X(-r)$$
We claim that $X(-R) \ne \emptyset$. If that is the case, we have $X(-r) \ne \emptyset$ and we can take a $p \in X(-r)$ and the ball centered at $p$ with radius $r$ lies completely within $F$. Since this is true for all $F$, we can conclude:
What's remain is to justify $X(-R) \ne \emptyset$. To see this, consider the function $$\varphi(x) = \inf\left\{ |y-x| : y \not\in X \right\}$$ It is easy to see $\varphi(x)$ is continuous over $X$. Since $X$ is compact, $\varphi(x)$ achieve maximum value $R_*$ at some $p_* \in X$. It is impossible for $R_* < R$ or we can move $p_*$ around to increase $\varphi(\cdot)$. This implies
$$R_* \ge R \implies p_* \in X(-R_*) \subset X(-R) \implies X(-R) \ne \emptyset$$
Notes
$\color{blue}{[1]}$, see $\S$ 1.7.6 of Santaló, L., & Kac, M. (2004). Integral Geometry and Geometric Probability (2nd ed., Cambridge Mathematical Library). doi:10.1017/CBO9780511617331
$\color{blue}{[2]}$ - We are using the result
This can be proved using the higher dimension generalization of Cauchy's surface area formula (see $\S 13.2$ of Santaló's book ) which relate the hyper-area $\mu_{d-1}(\cdots)$ with the angular average of projected areas.