It is not possible.
It is consistent with set theory without choice that $\mathbb R/\mathbb Q$ has strictly larger cardinality that $\mathbb R$. (This looks counter-intuitive, since $\mathbb R/\mathbb Q$ is a quotient.)
This is the case because, using a fact that goes back to Sierpiński (Sur une proposition qui entraîne l’existence des ensembles non mesurables, Fund. Math. 34, (1947), 157–162. MR0023318 (9,338i)), in any model of $\mathsf{ZF}$ where all sets of reals have the Baire property, it is not even possible to linearly order $\mathbb R/\mathbb Q$.
(Sierpiński argues in terms of Lebesgue measure. The argument in terms of the Baire property is analogous, and has the additional technical advantage of not requiring any discussion of consistency strength issues.)
A couple of years ago, Mike Oliver gave a nice talk on this topic (How to have more things by forgetting where you put them); he is not exactly using $\mathbb R/\mathbb Q$, but the arguments easily adapt. The point of the talk is precisely to give some intuition on why we expect the quotient to be "larger".
[Of course, in the presence of choice, the two sets have the same size. The argument above shows that the use of choice is unavoidable.]
Best Answer
Simply define $f(x) = \begin{cases}x + 1, \text{if } x \in \Bbb N;\\x, \text{else.}\end{cases}$, where $\Bbb N = \{0, 1, 2, \dots\}$ is the set of non-negative integers.
Then $f$ is a bijection from $\Bbb R$ to $\Bbb R\backslash \{0\}$.