The answer to your question is negative. A counterexample is given by the Morse function

$f:\mathbb{R}\to \mathbb{R}, \;\; f(x)=\sin(x). $

Using this example it is easy to produce higher dimensional counterexamples.

(The Morse functions with at most one critical point per level set were called *excellent* by R. Thom. They are also known as *stable functions*. They form an open and dense subset in the space of smooth functions on a manifold.)

**Re-Edit** (*Following Mark Grant's observation.*) If $f:\mathbb{R}\to\mathbb{R}$ is a polynomial

$$f(x)=x^n+\sum_{j=0}^{n-1} a_j x^j, $$

then your guess is correct. Namely for all but finitely many $a$'s the function $f_a(x):=f(x)-ax$ is excellent. Here is the argument.

Consider the real algebraic set $\newcommand{\bR}{\mathbb{R}}$

$$ Z=\lbrace (x,a) \in\bR^2;\;\; f'(x)=a\rbrace. $$

We have a natural projection $\alpha: Z\to \bR$, $(x,a)\mapsto a$, whose fibers are finite sets. This is a semialgebraic map so that there exist finitely many real numbers

$$A_0< A_1< ... < A_N$$

such that the map $\alpha$ is a $C^3$-covering map over each of the intervals $(-\infty, A_0)$, $(A_0,A_1)$, ... $(A_N,\infty)$.

Let $I$ be one of these intervals. The set

$$ Z(I):=\lbrace (x,a)\in Z;\;\;a\in I\rbrace $$

consists of the graphs of finitely many twice differentiable semialgebraic functions

$$ u_1,...,u_m: I\to\mathbb{R} $$

such that $ u_i(a) < u_{i+1}(a)$, $\forall a, i$. In other words, the set of critical points of $f_a$, $a\in I$ is $u_1(a),\dotsc , u_m(a)$.

I claim that for all but finitely many $a$'s in $I$ the function $f_a$ is excellent. I argue by contradiction. If this were not the case, then the semialgebraicity of the $u_i$'s implies that there exist an open interval $ J\subset I$ and pair of indices $(i,j)$, $i\neq j$, such that

$$ f_a( u_i(a) )= f_a( u_j(a) ), \;\;f'(u_i(a))=f'(u_j(a))=a,\;\;\forall a\in J.$$

The first equality can be rewritten as

$$ f(u_i)- f(u_j)= a(u_i-u_j). $$

Differentiating the above equality with respect to $a$ and recalling that $f'(u_i)=f'(u_j)=a$ we deduce

$$ a(u_i'-u_j')= (u_i-u_j)+ a(u_i'-u_j'). $$

This clearly implies

$$ u_i(a)=u_j(a),\;\;\forall a\in J. $$

*Contradiction!*

**Note.** The same is true for any polynomial $f$ in any number of variables. The proof in the general case is very similar to the one above in the one-dimensional case.