We have to distinguish between two algebraic structures with appropriate homomorphisms between them (in fact, they constitute categories)
$\bullet$ Rings (which I always assume to be unital) with homomorphisms of rings.
As always, a homomorphism of blupp preserves the whole structure of blupp, so in particular for blupp=Ring the unit is preserved by definition. This definition is (or should be ...) universally accepted.
$\bullet$ Rngs or non-unital rings with homomorphisms of rngs
Following the general principle, a homomorphism of rngs preserves the addition and the multiplication (and then also additive inverses and the zero), but not the unit because there is none.
There is a so-called forgetful functor from Rings to Rngs which forgets the unit. Notice that if $R$ is a ring, the underlying rng $|R|$ is a different object (and similarly for homomorphisms). This is usually ignored in textbooks and lectures, leading to utter confusions. Similar problems arise with other forgetful functors. These problems will end immediately when we take forgetful functors seriously. Unfortunately, some authors consider rings with unit but consider also non-unital homomorphisms between them, which doesn't make sense at all because this actually ignores the general principles of universal algebra, secretly applies the forgetful functor all the time, and doesn't take rings seriously. And some authors even don't consider $0$ as a ring because "it has no unit", sic! Of course the zero ring is a ring. It's quite cold these days, so books following this approach would make a warm fire ...
Anyway, now the question is easily answered:
Let $f : R \to R'$ be a homomorphism of rings (thus preserving the unit by definition). Then $f=0$ iff $R'=0$. It doesn't matter if $R$ is a field or not, the condition doesn't depend on $R$ at all.
On the other hand, there is always a zero homomorphism of the underlying rngs $0 : |R| \to |R'|$. Notice, again, that these are different objects than the rings themselves!
No, not at all. A homomorphism must take nilpotent elements to zero, since $\Bbb Z$ has no proper nilpotents. The matrix that’s all zero except for a $1$ in the upper right corner must thus be taken to $0$. Similarly for the matrix with $1$ in the lower left. But their sum squares to the identity matrix, so your homomorphism is zero. (There are much better abstract proofs.)
Best Answer
A ring homomorphism from a field to any ring is necessarily injective (assuming that ring homomorphisms map $1$ to $1$), because its kernel is a proper ideal and a field has only $\{0\}$ as proper ideal.
In this case it's also easy to verify it directly, because if $\phi(a+bi)$ is the null matrix, then necessarily $a=b=0$.
The verification that $\phi$ is a homomorphism consists in showing that $$ \phi(x+y)=\phi(x)+\phi(y),\qquad \phi(xy)=\phi(x)\phi(y),\qquad \phi(1)=\begin{bmatrix}1 & 0\\0&1\end{bmatrix} $$ for all $x,y\in\mathbb{C}$. For multiplication, consider $x=a+bi$ and $y=c+di$; then $xy=(ac-bd)+(ad+bc)i$, so $$ \phi(xy)=\begin{bmatrix}ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{bmatrix} $$ whereas $$ \phi(x)\phi(y)= \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} c & d \\ -d & c \end{bmatrix} $$ Can you finish doing the matrix product? The check for the addition should be carried on similarly.
Note. In the definition of $\phi$ it is implicitly assumed that $a,b\in\mathbb{R}$ and the same assumption is made in the check, also for $c$ and $d$.