Yes. Every positive map $f$ is self-adjoint: $f(x^*)=f(x)^*$ for every $x$. Hence, if $f(a)$=0 and $a=b+ic$ with $b,c$ self-adjoint, then $f(b)=0=f(c)$. Suppose for a contradiction that $b\neq0$. Then, for $b=b_+-b_-$ with positive $b_+,b_-$, one has $f(b_+)=f(b_-)$. The cp maps $g_{\pm}\colon{\mathbb C}\to A$, defined by $g_\pm(1)=b_\pm$, satisfy $f\circ g_+=f\circ g_-$.
For the epimorphism problem, let $f\colon A\to B$ be an epi cp contraction. Then, the unitization $f^{1}\colon A^1\to B^1$ is a unital cp map which is epi for ucp maps. Let $S$ be the norm closure of $f^{1}(A^{1})$ in $T:=B^{1}$. We will prove that if $S\subset T$ is proper inclusion of operator systems (those are unital self-adjoint closed subspaces of a $\mathrm{C}^*$-algebra), there is a states $g,h$ on $T$ such that $g|_S=h|_S$ but $g\neq h$. By the Hahn--Banach theorem, there is a non-zero bounded linear functional $\phi$ on $T$ such that $\phi|_S=0$. By the GNS construction for a bounded linear functional (of an ambient $\mathrm{C}^*$-algebra), we may assume that $\phi(x)=\langle x\eta,\xi\rangle$ for some realization $T\subset B(H)$ and $\xi,\eta\in H$. Notice that $\xi\perp\eta$ because $1\in S$. Now we consider the states $g'$ and $h'$ on $T$ defined by $2g'(x)=\langle x\xi,\xi\rangle+\langle x\eta,\eta\rangle$ and by $2h'(x)=\langle x(\xi+\eta),\xi+\eta\rangle$. They coincide on $\Phi(S)$ but not on $\Phi(T)$. Hence, the states $g=g'\circ\Phi$ and $h=h'\circ\Phi$ on $T$ will do.
