The answer is no, and there are plenty of counterexamples. Note that simplicial sets are not relevent here; one can cook up examples with spaces and take their total singular complexes.


For example, there are high dimensional knots $K: S^n \to S^{n+2}$ (i.e., smooth embeddings with $n > 1$) such that the complement $X = S^{n+2} - K(S^n)$ has $\pi_1(X) = \Bbb Z$. A generator is represented by a map $X \to S^1$ which is a both a $\pi_1$- and a homology isomorphism.   This will give examples with the exception of your condition on $\pi_2$.


To get the $\pi_2$ condition on the above consider the subclass of those knots such that $n = 2k+1$ is odd and $\pi_j(X) \cong \pi_j(S^1)$ for $j\le k$ and $\pi_{k+1}(X) \ne 0$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing).  The classification was announced by Kearton in the paper 

*Classification of simple knots by Blanchfield duality.* Bull. Amer. Math. Soc. **79** (1973), 952–955