Let $\mathrm{Sym}^2(\mathbb{R}^n)$ be the $\binom{n+1}{2}$ dimensional vector space of $n \times n$ symmetric matrices and let $P \subset \mathrm{Sym}^2(\mathbb{R}^n)$ be the cone of positive semidefinite matrices. A spectrahedron is the intersection of $P$ with an affine linear space $L$. One reasonable interpretation of the "almost always" in your question is "except for a measure $0$ subset of the Grassmannian of $d$-dimensional affine subspaces of $\mathrm{Sym}^2(\mathbb{R}^n)$. With that interpretation, the answer is "yes" for $d \leq 2$, but "no" for $d \geq 3$.

The boundary $\partial (P \cap L)$ is $\partial P \cap L$. The boundary $\partial P$ is singular along the $\binom{n+1}{2}-3$ dimensional space $Q$ of positive semidefinite matrices of rank $\leq n-2$, and smooth at the matrices of rank $n-1$. If $L$ passes through $Q$ then $\partial P \cap L$ is almost surely singular at $Q \cap L$; if $L$ misses $Q$ then $L$ is likely to be singular.

Basic dimension counting (like the proof of Bertini's theorem, though this is in the real setting) shows that, since $Q$ has codimension $3$, a positive volume set of $L$ planes meets $Q$ for $d \geq 3$, but only (using Sard's theorem) a measure $0$ set for $d \leq 2$.