I think "yes". In the following let $\alpha$ and $\beta$ denote the Kuratowski and the Hausdorff measure of non-compactness, respectively. Note that
$$
\beta(A) \le \alpha(A) \le 2 \beta(A)
$$
for each bounded subset $A$ of a Banach space $E$. The following example is in $E:=c_0(\mathbb{N},\mathbb{R})$ (endowed with the maximum norm $\|\cdot\|$). In this space we have the following representation of $\beta$: For each bounded set $A \subseteq E$:
$$
\beta(A)= \lim_{n \to \infty} (\sup_{x \in A} (\max \{|x_k|: k \ge n\})),
$$
see Banas J., Goebel K. Measures of noncompactness in Banach spaces, 6.1.
Let $\varphi: \mathbb{R} \to \mathbb{R}$ be the function
$$
\varphi(t)=\frac{t}{4} \sin(1/t) ~ (t \not=0), \quad \varphi(0)=0,
$$
and note that $|\varphi(t)| \le |t|/4$.
Let $K$ denote the closed unit ball in $E$ and set $f:K \to K$ as
$$
f(x)=f((x_k)_{k=1}^\infty) :=(\varphi(x_k))_{k=1}^\infty.
$$
Now, for each set $A \subseteq K$ we have
$$
\beta(f(A)) = \lim_{n \to \infty} \sup_{y \in f(A)} \max \{|y_k|: k \ge n\}
= \lim_{n \to \infty} \sup_{x \in A} \max \{|\varphi(x_k)|: k \ge n\} \le \frac{1}{4} \beta(A),
$$
thus
$$
\alpha(f(A)) \le \frac{1}{2} \beta(A) \le \frac{1}{2}\alpha(A).
$$
Now we show that $f$ is not the sum of a contraction (even a Lipschitz map) and a compact map:
Assume by contradiction that there are functions $g, h:K \to E$ such that
1.) $\forall x \in K$: $f(x)=g(x)+h(x)$,
2.) $\exists L \ge 0$ $\forall x,y \in K$: $\|g(x)-g(y)\| \le L\|x-y\|$,
3.) $\overline {h(K)}$ is compact.
Since the derivative of $\varphi$ is unbounded (in both directions) near $0$ we find $-1 \le \eta < \xi \le 1$ such that
$$
(\ast) \quad \varphi(\xi)-\varphi(\eta) > L (\xi-\eta).
$$
For $m \in \mathbb{N}$ set $e^{(m)}:=(\delta_{k,m})=(0,\dots,0,1,0,\dots)$, and
$$
x^{(m)}:= \xi e^{(m)}, ~~ y^{(m)}:= \eta e^{(m)}, ~~ u^{(m)}:= g(x^{(m)})-g(y^{(m)}), ~~ v^{(m)}:= h(x^{(m)})-h(y^{(m)})
$$
(note that $x^{(m)},y^{(m)} \in K$).
By 1.) and 2.) we have for each $m$:
$$
(\ast\ast) ~~ \varphi(\xi)-\varphi(\eta)=u^{(m)}_m + v^{(m)}_m \le L\|x^{(m)}-y^{(m)}\| + v^{(m)}_m = L(\xi-\eta)+ v^{(m)}_m.
$$
Next, by 3.), the sequence $(v^{(m)})$ has a convergent subsequence $(v^{(m_j)})$ with limit $w=(w_k) \in E$, say. We have
$$
|v^{(m_j)}_{m_j}| \le |v^{(m_j)}_{m_j}-w_{m_j}| + |w_{m_j}|\le \|v^{(m_j)}-w\| + |w_{m_j}| \to 0 \quad (j \to \infty).
$$
Now $(\ast\ast)$ with $m=m_j$ and $j \to \infty$ yields $\varphi(\xi)-\varphi(\eta) \le L (\xi-\eta)$,
a contradiction to $(\ast)$.
Hope everything is correct, please check.
