I am writing my comments as a solution. Assume that $k$ has characteristic $0$, for simplicity. First I will make precise one interpretation of "vanishing monodromy around the discriminant". Denote by $\Delta$ the complement of $B^{\text{sm}}$ in $B$. For simplicity, assume that $\Delta$ has pure codimension $1$ in $B$ (components of $\Delta$ that have codimension $\geq 2$ have no "local monodromy" by the Purity Theorem for fundamental groups). For every irreducible component $\Delta_i$ of $\Delta$, for the generic point $\eta_i$ of $\Delta_i$, for the strict Henselization of the local ring $$R_i:=\mathcal{O}^{\text{sh}}_{B,\eta_i},$$ the (profinite) Galois group of the fraction field of this local ring, $$\widehat{\pi}^{\text{inertia}}_1(B^\text{sm};\overline{\eta}_i):= \text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i)),$$ equals the profinite completion of $\mathbb{Z}$ (this is essentially the Newton-Puiseux Theorem), i.e., it is procyclic. If $k$ equals $\mathbb{C}$, for a general $\mathbb{C}$-point $p_i$ of $\Delta_i$, for a transverse slice $N$ to $\Delta_i$ at $p_i$, for the punctured neighborhood $N^*=N\setminus\{p\}$, this procyclic group corresponds to the profinite completion of $\pi_1(N^*)\cong \mathbb{Z}$.
Denote by $X_{\overline{\eta}_i}$ the fiber product of $X\to B$ and $\text{Spec}(\overline{\text{Frac}(R_i)}) \to B.$ Then the profinite Galois group $\text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i))$ acts continuously on the etale cohomology of the $\overline{\text{Frac}(R_i)}$-scheme $X_{\overline{\eta}_i}$. If there is a modification of $X$ that makes $f$ smooth over $\eta_i$, then this action is trivial. This is what is usually meant by vanishing monodromy on cohomology of the fiber. If $k$ equals $\mathbb{C}$, then (via the usual comparison theorems) this is equivalent to saying that the linear representation of $\pi_1(N^*,q)$ on the singular cohomology of the fiber $X_q$ over $q\in N^*$ is a trivial representation.
Vanishing of monodromy on cohomology of the fiber is not sufficient for the existence of a modification. Here is one example. Let $B$ be the projective plane $\text{Proj}\ k[r,s,t]$. Let $\mathbb{P}^2_k$ be the projective plane $\text{Proj}\ k[u,v,w]$. Let $X$ be the closed subscheme of $B\times \mathbb{P}^2_k$ with defining equation, $$f=u^2r+v^2s+w^2t.$$ Then $X$ is smooth, and the projection $f$ to $B$ is flat. Moreover, the fiber $X_{\overline{\eta}_i}$ is just $\mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$. The étale cohomology of this is straightforward: $H^q(X_{\overline{\eta}_i},\widehat{\mathbb{Z}}_\ell)$ equals $\widehat{\mathbb{Z}}_\ell$ for $q=0$, it is zero for $q=1$, and it equals $\widehat{\mathbb{Z}}_\ell(-1)$ for $q=2$. Of course the action of monodromy on $H^0$ and $H^1$ is trivial. For $q=2$, observe that the first Chern class of the relative dualizing sheaf is a nonzero element that is preserved under monodromy. Since $H^2$ is procyclic and the action is continuous, it follows that the monodromy also acts trivially on $H^2$.
However, there is no modification of $X$ that makes $f$ smooth over the generic points $\eta_i$ of the three coordinate axes in $B$. If there were such a modification, then by Zariski glueing, there exists a modification of $X$ that is smooth over a dense Zariski open $B^o$ in $B$ containing both $B^{\text{sm}}$ and containing the generic point $\eta_i$ of each irreducible component $\Delta_i$. Thus, the complement of $B^o$ is a closed subset of $B$ that has codimension $\geq 2$. By the Purity Theorem for Brauer groups, Théorème 6.1 of the following, the restriction map from the Brauer group of $B$ to the Brauer group of $B^o$ is an isomorphism.
MR0244271 (39 #5586c)
Grothendieck, Alexander
Le groupe de Brauer. III. Exemples et compléments.
(French) Dix exposés sur la cohomologie des schémas, 88–188,
Adv. Stud. Pure Math., 3,
North-Holland, Amsterdam, 1968.
The Brauer group of $B$ is zero. This implies that the Brauer class of $X^{\text{sm}}/B^{\text{sm}}$ is zero, and thus there exists an element of the Picard group of $X^{\text{sm}}$ whose base change to $X_{\overline{\eta}_i} \cong \mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$ is a generator of the Picard group. Since $X$ is smooth, every divisor class on $X^{\text{sm}}$ is the restriction of a divisor class on $X$. The projection from $X$ to $\mathbb{P}^2_k$ is the projective bundle associated to a locally free sheaf of rank $2$, from which it is straightforward to compute the Picard group of $X$. There is no divisor class on $X$ whose base change to $X_{\overline{\eta}_i}$ is a generator of the Picard group. This contradiction proves that there is no modification of $X$ that is smooth over a generic point $\eta_i$ of $D_i$.
