An $R$-linear homomorphism has to take $1\mapsto 1$ and then be $R$-linear, so that $r\mapsto r$. All $R$-linear functions leaving $R$ are therefore completely determined.
Consider $R=\Bbb C$: Notice that evaluation mappings out of $\Bbb C[x]$ produce distinct mappings to $\Bbb C$.
You started the context in Alg(R) and then asked about injectivity and projectivity, but we most often talk about injectivity and projectivity in the category of modules. In the context of modules, $R$ is as you noted always a free module over itself, but it is not always injective module. Such rings $R$ are called self-injective rings.
In a principal ideal domain, like $\Bbb Z$, each ideal is module isomorphic to $\Bbb Z$. If $\Bbb Z$ is injective, so would these submodules be injective. But injective submodules are always summands of their supermodules. Can you see why this isn't possible in a domain like $\Bbb Z$?
If you really want to talk about free, projective and injective objects in the category of Alg(R), let me know.