Does one real radical root imply they all are? Is there an example of an irreducible polynomial $f(x) \in \mathbb{Q}[x]$ with a real root expressible in terms of real radicals and another real root not expressible in terms of real radicals? 
 A: The answer to the question is yes (so the answer to the title is no) and I will give an example later.
Let me first recall a couple of results. The first one is the following, that can be found in [Cox, Galois Theory, Theorem 8.6.5].

Theorem 1. Let $F$ be a subfield of $\mathbb{R}$ and let $f \in F[x]$ be an irreducible polynomial with splitting field $F \subset L \subset \mathbb{R}$. Then the following conditions are equivalent.
(1) Some root of $f$ is expressible by real radicals over $F$.
(2) All roots of $f$ are expressible by real radicals over $F$ in which only square roots appear.
(3) $[L:F]$ is a power of $2$.

So, if $f$ splits completely over $\mathbb{R}$, the existence of a root expressible by real radicals forces all the roots to be so.
On the other hand, when $f$ does not split completely in $\mathbb{R}$ this is no longer true. Let us state our second result, that can be found in
[A. Loewy, Über die Reduktion algebraischer Gleichungen durch Adjunktion insbesondere reeller Radikale, Math. Zeitschr. 15, 261-273 (1922)],
see also Cox book, Theorem 8.6.12.

Theorem 2. Let $F$ be a subfield of $\mathbb{R}$ and $f \in F[x]$ irreducible of degree $2^mn$, with $n$ odd. Then $f$ has at most $2^m$ roots expressible by real radicals over $F$.

In particular, when $f \in \mathbb{Q}[x]$ is irreducible and of odd degree, Theorem 2 implies that at most one root of $f$ is expressible by real radicals. Note that if the degree is $3$ then Theorem 2 is consistent with Cardano's formulas, and if the degree is a power of $2$ then it is consistent with Theorem 1.
Finally, let us give the following example answering the question, that can be found in Loewy's paper quoted above, page 272. Let us consider the polynomial $$x^6+6x^4-234x^2-54x-3 =(x^3+(3+9 \sqrt{3})x+ \sqrt{3})(x^3+(3-9 \sqrt{3})x- \sqrt{3}).$$
It is irreducible over $\mathbb{Q}$ by Eisenstein's criterion and it has one real root expressible by real radicals, three real roots not expressible by real radicals and two complex roots.
A: WARNING: Wrong answer follows, as per comments. Kept here to deter others from making the same mistake. 
Perhaps $$x_1=\sqrt{2+2^{1/4}}+\sqrt{2-2^{1/4}},\quad x_2=\sqrt{2+i2^{1/4}}+\sqrt{2-i2^{1/4}}$$ are real numbers, solutions of $\bigl((x^2-4)^2-16\bigr)^2=32$, one expressed using real radicals, the other, not. 
