Charge in Yukawa interaction The Yukawa interaction Lagrangian is:
$$
\mathcal{L}_{int}=-g\bar{\psi}(x)\psi(x)\phi(x)+h.c.
$$
Is it possible to assign "charge" to any of the particles, like an electric charge? My first instinct would be no, but I am not sure if I can justify it - is it possibly because the strength of the interaction is dependent on three particles, rather than an even amount of them?
On the other hand, I know that the Yukawa interaction is supposed to be an approximation of the strong force - so perhaps a version of the color charge can be assigned?
 A: No! A straight $\pm$ charge is not possible, as the (Abelian) Yukawa force can never be repulsive. There is a generic theorem on charge conjugation to the effect that if the spin of the exchange particle (here, the scalar) is even, the force can only be attractive, but if it is a vector (like the photon of EM), it can be either attractive or repulsive, depending on the (abelian) charge,  $\pm e$.
Nevertheless, (so,   "& yes"!), you may complicate matters in the non-Abelian case, by introducing several intermediating particles. For instance, the actual interaction with pseudoscalar pions is $-g\bar\psi \vec \tau \gamma_5 \psi  \cdot \vec \pi $,  which may and does  complicate the landscape, as isosinglets are split from isotriplets. So the answer to your question is "no & arguably yes", the latter a wink to contrarians and fussbudgets.
(PS The charge above is isospin, not color; the pions are colorless...)
A: Cosmas Zachos provided an excellent answer in the context of QCD. Let me add a bit more insight on the electroweak side of the story.
In the context of Higgs mechanism, the Yukawa interaction is
$$
\mathcal{L}_{int}=-g\bar{L}(x)\phi(x)e_R(x)+h.c.
$$
where ${L}(x)$ is the left-handed (electron  + neutrino) doublet $\begin{pmatrix} \nu_L(x) \\ e_L(x) \end{pmatrix}$, $e_R(x)$ is the right-handed electron singlet, and Higgs scalar $\phi(x)$ is isospin doublet $\begin{pmatrix} \phi_{1}(x) \\ \phi_{2}(x) \end{pmatrix}$. Under the electromagnetic charge gauge transformation, the individual fields transform as
$$
e_R(x) \rightarrow e^{-i\theta}e_R(x) \quad :\quad q = -1\\
e_L(x) \rightarrow e^{-i\theta}e_L(x) \quad :\quad q = -1\\
\nu_L(x) \rightarrow \nu_L(x) \quad \quad :\quad q = 0\\
\phi(x) \rightarrow e^{i\theta(\tau_3+\frac{1}{2})}\phi(x) \quad :\quad q = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
where $\tau_3$ is the isospin operator. And the Lagrangian is guaranteed to be gauge invariant.
As you can see, the Higgs doublet $\phi(x)$ is in general NOT electric  charge zero! It's the only vacume expectation value of the Higgs doublet
$$
\phi(x) = \begin{pmatrix} 0 \\ \upsilon \end{pmatrix}
$$
that is charge zero, since the nonzero VEV is the isospin down component of the Higgs doublet.
According to the Higgs folklore, the charge $q=1$ component of the Higgs doublet was devoured by the $W^\pm$ gauge fields, and resultantly $W^\pm$ gained weight after the feast, albeit it remained a mystery  as to how much a weight was put on by $W^\pm$.
