Suppose that $A$ has discrete spectrum consisting of eigenvalues with finite multiplicities

$$ 0<\lambda_1 < \lambda_2<\cdots $$

with $\lambda_n\to\infty$ as $n\to \infty$. Denote by $(\psi_n)$ an orthonormal basis consisting of e-vectors of $A$. In this basis $A$ is an (infinite) diagonal matrix $\DeclareMathOperator{\diag}{diag}$

$$ A=\diag(\lambda_1,\lambda_2,\dotsc, ...).$$

Then $\newcommand{\ve}{\varepsilon}$

$$f(A+\ve I)=\diag( f(\lambda_1+\ve), f(\lambda_2+\ve),\dotsc),..). $$

Assume that $\newcommand{\bR}{\mathbb{R}}$ $f:\bR\to\bR$ is $C^2$. Then

$$ f(\lambda_n+\ve I)=f(\lambda_n)+\ve f'(\lambda_n)\ve+ \frac{1}{2}f''(\xi_n)\ve^2,\;\;\xi_n\in (\lambda_n,\lambda_n+\ve). $$

Suppose more concretely that $f(x)=x^3$. The above shows that $f(A+\ve I)-f(A)$ is not bounded.

This may not be as surprising since $A^3$ is not a bounded operators. Here is a more interesting examples.

Suppose for simplicity that $\lambda_n=n$ and $f$ is a $C^2$ function such that for $|x-n|<0.1$ we have $f(x)=n^2\cos (x-n)-n^2+1$. Note that

$$f(\lambda_n)=1,\;\;\forall n $$

so $f(A)=I$. Consider the bounded operator

$$ B=\diag( 1^{-1/2}, 2^{-1/2},\dotsc, n^{-1/2},\dotsc, ). $$

Then

$$f(A+\ve B)-f(A)=\diag(\dotsc, f(n+\ve n^{-1/2})-f(n),\dotsc), $$

and we observe that, if $\ve<0.1$, then

$$ f(n+\ve n^{-1/2})-f(n) =n^2\cos\ve n^{-1/2}. $$

We deduce that $f(A+\ve B)-f(A)$ is not bounded.

**Remark** (a) Here is a possible reformulations. There are several natural topologies on the space of closed selfadjoint operators; see this paper. One of them, called *gap topology* in the above paper is defined by a certain metric $\gamma$ (distance between the graphs) and has the property

$$\gamma(A_n,A)\to 0 \,\Longleftrightarrow\; \Vert f(A_n)- f(A)\Vert\to 0,\;\;\forall f\in C_0(\bR), $$

where $C_0(\bR)$ denotes the space of continuous functions $f:\bR\to\bR$ such that

$$\lim_{t\to\pm\infty} f(t)=0. $$

It may be the case that if $f\in C^2(\bR)$ is such that $f,f', f''\in C_0(\bR)$ then a conclusion of the type you've formulated could be true.

(b) Let me mention a closely related question. The set of closed selfadjoint *Fredholm* operators on $H$, equipped with the above gap topology can be organized as a Banach manifold. More precisely it is an open dense subset $\newcommand{\eO}{\mathscr{O}}$ $\eO$ of $\DeclareMathOperator{\Lag}{Lag}$ $\Lag(H\oplus L)$ Grassmannian of Lagrangian subspaces of $H\oplus H$; see this paper or this paper for more details. The above remark shows that any $f\in C_0(\bR)$ defines a continuous function $\hat{f}:\eO\to\bR$, $A\mapsto f(A)$ and I am $52$% sure that it extends to $\Lag(H\oplus H)$. Is it true that if $f\in C_0(\bR)$ is such that $f'\in C_0(\bR)$ that $\hat{f}$ is a $C^1$-function on $\eO$?