Yes, your example (pieced together from the original post and the comments) $\begin{bmatrix}F_2&F_2\\0&0\end{bmatrix}$ is a noncommutative rng without identity of order $4$.
Since $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ acts as an identity on the left, it would have to be equal to any candidate two-sided identity, if it existed, but $\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq \begin{bmatrix}0&1\\0&0\end{bmatrix}$
Obviously you can't have such a ring with $1$ or $2$ elements.
If such a ring had $3$ elements, say $\{0,a,b\}$, all distinct, then $a+b$ must be somewhere in this set. $a+b\in \{a,b\}$ creates a contradiction, so $a+b=0$ necessarily, so that $b=-a$. But multiplication in $\{-a,0,a\}$ necessarily commutes.
What you gave is actually a nice representation for one of my favorite examples of semigroup rings. You take the semigroup $S=\{a,b\}$ with multiplication defined by $a^2=ba=a$ and $b^2=ab=b$, and make the semigroup ring $F_2[S]$. I think this is the same ring as if we had taken $a=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $b=\begin{bmatrix}1&1\\0&0\end{bmatrix}$.
As shown here, since projectives are injectives, you can always say that a module over a QF ring having finite projective dimension is already projective. Therefore the projective dimension of a module over a QF ring is either $0$ or $\infty$.
So adding your second condition amounts to every module being projective, and thus you have a semisimple ring.
Best Answer
In an earlier version of this post I caused an accident by giving a wrong answer. Thanks to those who pointed out the error! Here is a little update:
The ring $M_2(\Bbb F_2)$ of $2 \times 2$-matrices with entries in $\Bbb F_2$ is a non-commutative ring with 16 elements, because $$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$ It has a subring of order $8$, namely the upper triangular matrices which are non-commutative by the example above.
We show that $8$ is minimal:
Let $R$ be a finite ring with unity with $n$ elements.
We need two preliminaries:
1.) If the additive group is cyclic, then $R$ is commutative.
Proof: If the additive group of $R$ is cyclic, we can choose $1$ as a generator: If we have $0 = 1+\dots+1 = m \cdot 1$ for some $m\in \Bbb N$, then $0=(m\cdot 1) \cdot g = m\cdot(1\cdot g)=m\cdot g$ so the additive order of $1$ is maximal. Thus, the multiplication table of $R$ is determined by $1 \cdot 1 = 1$, showing $R \cong \Bbb Z/n \Bbb Z$ and $R$ is commutative.
2.) All rings of order $4$ are commutative. Proof: As a general result, all rings with order equal to a squared prime are commutative: Ring of order $p^2$ is commutative.
Thus any ring of order $1,2,3,5,6,7$ is ruled out by 1.) using the Sylow-theorems and $4$ is ruled out by 2.).
So $8$ is the minimal cardinality a non-commutative ring can have.