No, even if there is a finite number of feasible rank-1 matrices, the feasible region of an SDP does not have to be polyhedral.
A spectrahedron you see all the time in applications is $S_n = \{X: X \succeq 0, X_{11} = \ldots = X_{nn} = 1\}$, i.e. the set of Gram matrices of $n$ unit vectors. This is, for example, the feasible region for the Goemans-Williamson SDP relaxation for MaxCut. There can be no more than $2^n$ rank-1 matrices in $S_n$, because $xx^T \in S_n$ implies $x_i^2 = 1$ for all $i$, and therefore $x \in \{-1, 1\}^n$.
Now let's look at $S_3$. Write
$$
X = \left(
\begin{array}{ccc}
1 & x & y\\
x & 1 & z\\
y & z & 1
\end{array}
\right)
$$
By Sylvester's criterion, $X \succeq 0$ if and only if all principal minors are non-negative. This gives the following inequalities:
$$
\begin{align}
x^2, y^2, z^2 &\leq 1\\
x^2 + y^2 + z^2 &\leq 1 + 2xyz
\end{align}
$$
The first three inequalities come from writing the 2-by-2 minors, and the last comes from writing the determinant of $X$.
It's now easy to see this set is not polyhedral. For example, let the set $T$ be the projection of $S_3$ onto the free variables $x, y, z$, and consider $U = T \cap \{(x, y, z): z = 0\}$. Polyhedral sets remain polyhedral after orthogonal projection and intersection with halfspaces, so if $S_3$ is polyhedral, then $U$ is as well. But $U = \{(x, y, 0): x^2 + y^2 \leq 1\}$ is a disc.
In fact there is also a direct geometric argument that $U$ is a disc. If $X$ is the Gram matrix of the vectors $u, v, w$, then setting $z = 0$ means $v \perp w$, and $(x, y)$ are the coordinates of the projection of $u$ onto the plane spanned by $v$ and $w$, expressed in the orthonormal basis given by $v$ and $w$. Since $u$ can be any unit vector, $(x,y)$ can be any vector of length at most $1$.
For illustration, here is the set $T$:
And here you can see that $U$ is a disc:
