This is not a complete answer to the original question, since the original question has a rather open-ended phrasing; but I think it addresses Diego's main points, and shows that even quite well-behaved cp maps won't have the desired properties.
First of all: if we restrict attention to cases where either $A$ or $B$ is commutative, then every positive linear map $A\to B$ is automaticaly completely positive: see Lemma 5.1.4 and Theorem 5.1.5 in the book of Effros--Ruan, for instance. To keep things simple I will take both $A$ and $B$ to be commutative ${\rm C^*}$-algebras; I'm sure one can create more fancy examples by modifying the constructions below.
Q1: "Are injective norm-one cp maps isometric?"
Counterexample: take $A=B = \ell_\infty^2$ (viewed as $C(\{0,1\})$) and take the linear map $T:A\to B$ defined by $T(x,y) = (x, y/3)$.
Q2: "Are ucp maps isometric?"
This counterexample requires a bit more work than I originally thought, so it seems to deserve writing down as an answer rather than merely a comment. We work with $A=B=C[0,1]$ and consider positive, unit-preserving operators $T:C[0,1]\to C[0,1]$ of the form $$ (Tf)(x) = \int_0^1 K(x,t) f(t)\,dt $$ for a suitably chosen positive kernel function $K$. The condition $T1 = 1$ translates into requiring $\int_0^1 K(x,t)\,dt=1$ for all $x$, so that each "row" of $K$ should have total mass 1, and then on the analysts' general principle that "to make things go wrong, get things to pile up more and more" it is natural to try an Ansatz like $$ K(x,t) = \begin{cases} x^{-1} & \hbox{ if $0\leq t\leq x$} \\ 0 & \hbox{if $x< t \leq 1$} \end{cases} $$ With this choice, our operator $T$ becomes $$ (Tf)(x) = \frac{1}{x} \int_0^x f(t)\,dt $$ (sometimes known as the Hardy operator, if I recall correctly) and then we can see directly that $T$ is positive and unit-preserving.
On the other hand, because $T$ is averaging the behaviour of a given $f$ over the initial parts of the interval $[0,1]$, one intuitively feels that $T$ has a good chance of shrinking the norms of functions which are small near $0$ and big near $1$, which will mean that $T$ can take elements of norm $1$ to elements with small norm. Therefore we consider $Tf_n$, where $f_n(t)=t^n$, and calculate that $$ (Tf_n)(x) = \frac{1}{x} \int_0^x t^n\,dt = \frac{1}{n+1} x^n $$ so that each $f_n$ is in fact an eigenvector of $T$ with eigenvalue $(n+1)^{-1}$. In particular, $T$ is not bounded below.
To conclude that $T$ does not have closed range, it only remains to note that $T$ is injective. This is immediately plausible as we are basically integrating and then dividing by a weight, and taking the indefinite integral of a non-zero function should give a non-zero function. More precisely: suppose $f\in C[0,1]$ and $Tf=0$. Then $\int_0^x f(t)\,dt=0$. So by the fundamental theorem of calculus $f=0$.d