Here's a class of counterexamples for the pointed homotopy category of connected CW complexes (so even this restriction does not save you). Let $hCW_{\ast}$ denote this category, and let $\pi_{\bullet} : hCW_{\ast} \to \text{Set}_{\bullet}$ denote the functor taking a pointed CW complex to its homotopy groups $\pi_n$. By Whitehead's theorem, $\pi_{\bullet}$ is a conservative functor: it reflects isomorphisms in the sense that if a morphism $f$ has the property that $\pi_{\bullet}(f)$ is an isomorphism, then $f$ must be an isomorphism. Because $\pi_{\bullet}$ consists of a sequence of representable functors, it also preserves any limits that exist in $hCW_{\ast}$. In particular, $hCW_{\ast}$ has, and $\pi_{\bullet}$ preserves, finite products.

On the other hand, $\pi_{\bullet}$ is not a faithful functor: if $\pi_{\bullet}(f) = \pi_{\bullet}(g)$ it does not follow that $f = g$ in the homotopy category (so Whitehead's theorem for maps fails). On the third hand:

**Lemma 1:** If $C, D$ are categories with equalizers and $F : C \to D$ is a conservative functor which preserves equalizers, then $F$ is faithful.

*Proof.* Two parallel morphisms $f, g : c \to d$ in $C$ are equal iff their equalizer, as a map to $c$, is an isomorphism. Since $F$ preserves equalizers and is conservative, it also reflects equalizers, so if $F(f)$ and $F(g)$ have an equalizer in $D$, then $f$ and $g$ have an equalizer in $C$. Moreover, if $F(f) = F(g)$, then $F(f)$ and $F(g)$ have equalizer an isomorphism to $F(c)$, and since $F$ is conservative, $f$ and $g$ must also have equalizer an isomorphism to $c$, and so must be equal. $\Box$

**Lemma 2:** If $C$ has finite products and pullbacks, then it has equalizers. If $F$ preserves finite products and pullbacks, then it preserves equalizers.

*Proof.* Suppose $f, g : c \to d$ are two parallel morphisms. Then their equalizer is equivalently the pullback of the diagram $c \xrightarrow{(\text{id}_c, f)} c \times d \xleftarrow {(\text{id}_c, g)} c$. In other words, it's given by the intersection of their "graphs" in $c \times d$. $\Box$

**Corollary:** $hCW_{\ast}$ does not have pullbacks.

This argument is constructive in the sense that any example of a pair of parallel maps $f, g : X \to Y$ in $hCW_{\ast}$ such that $\pi_{\bullet}(f) = \pi_{\bullet}(g)$ but $f \neq g$ produces an example of a pullback that does not exist in $hCW_{\ast}$, namely the pullback of the diagram

$$X \xrightarrow{(\text{id}_X, f)} X \times Y \xleftarrow{(\text{id}_X, g)} X.$$

There are many examples of such maps; for example, we can take $X$ and $Y$ to have homotopy groups in disjoint degrees. The simplest examples have $X = BG, Y = B^2 A$ for some group $G$ and some abelian group $A$; then homotopy classes of maps $X \to Y$ are given by group cohomology classes in $H^2(G, A)$, or equivalently by central extensions of $G$ by $A$. But $\pi_{\bullet}(f) = 0$ for any map $f : X \to Y$.