Anton's answer prompted me to pull out Rudin's Functional Analysis. Exercise 23 of Chapter 3 is to exhibit the Gelfand-Pettis integral of a continuous function with values in a Frechet space as a strong limit of "Riemann sums" (looks slightly stronger than the Bochner integral, in fact). I couldn't make sense of how to use the hint, so I decided to see what I could come up with. It's been a long time since I've done this sort of thing, so I may have done something stupid.

We're going to assume $X$ is compact, $\mu$ is a probability Borel measure, $f$ is continuous, $E$ is locally convex and satisfies the technical criterion for the existence of Gelfand-Pettis integrals (closure of the convex hull of the image is compact). Then $f(X)$ is compact. If we also assume either $E$ is a metric space or $X$ is separable, then $f(X)$ is separable. Take a countable dense subset $e_i$ of $f(X)$. For a finite subset $B$ of seminorms and $\epsilon>0$, let $V(B,\epsilon,e_i)=\{e:p(e-e_i)<\epsilon\}$. These form a basis for the topology of $E$ (and they are convex and balanced).

First, we're going to construct a sequence of simple functions that converges to $f$, mimicking the example from measure theory.

Since $f(X)$ is compact, we can form a cover with finitely many of the $V(B,\epsilon,e_i)$, call them $A_i$. Let $E_i=A_i-\bigcup_{j=1}^{i-1}A_j$. Set $X_i=f^{-1}(E_i)$, so the $X_i$ are a disjoint partition of $X$.

Set $g_{B,\epsilon}(x)=\sum_i 1_{X_i}(x)e_i$ (where $i$ ranges over a finite set).
Consider $f(x)-g_{B,\epsilon}(x)$. If $x\in X_i$, $f(x)\in E_i$, so for all $p\in B$, $p\big(f(x)-g_{B,\epsilon}(x)\big)=p\big(f(x)-e_i\big)<\epsilon$. Convergence is easy from this.

Now we will show that $\int_X g_{B,\epsilon}\ d\mu=\sum_i \mu(X_i)e_i$ converges. We "cheat" by showing it converges to the Gelfand-Pettis integral of $f$ (instead of e.g. showing that it is Cauchy). So consider
$$\int_X f\ d\mu- \int_X g_{B,\epsilon}\ d\mu=\int_X f- g_{B,\epsilon}\ d\mu=\sum_i\int_{E_i}f-e_i\ d\mu$$

On $E_i$, $f-e_i$ lies in $V(B,\epsilon)$, so $\int_{E_i}f-e_i\ d\mu$ is in the closure of $\mu(E_i)\cdot V(B,\epsilon)\subset \overline{V(B,\epsilon)}$ (since $V(B,\epsilon)$ is balanced) (this uses the estimate that the Gelfand-Pettis integral lies in the closure of the convex hull of the image, times the measure of the space). So $\sum_i\int_{E_i}f-e_i\ d\mu\subset \sum_i \overline{V(B,\epsilon)}\subset \overline{\sum_i V(B,\epsilon)}$.

By some basic tvs theorems, since $V(B,\epsilon)$ forms a basis for the topology, we can choose $V(B,\epsilon)$ so that $\overline{\sum_i V(B,\epsilon)}$ is inside any given neighborhood of $0$ (see p.10-11 of Rudin).

So Bochner integrals of continuous functions exist whenever Gelfand-Pettis integrals exist (plus a separability assumption), unless I made a mistake. Even though I left it in, I don't think that I used separability in an essential way (I could have just chosen an uncountable dense subset $e_i$ and proceeded). That requirement for strong measurability (almost-separably valued) may be an artifact of using sequences instead of nets.