The answer is yes for continuous functions $f$ on continuous curves $C$. The reason for this is that the line integral (with respect to arc length) satisfies the inequality
$$ \min(f) \cdot L_C \leq \int_C f(x_1, \ldots x_n) \,ds \leq \max(f) \cdot L_C $$
which can be rearranged as
$$ \min(f) \leq \cfrac{1}{L_C} \int_C f(x_1, \ldots x_n) \,ds \leq \max(f) $$
We know that the minimum and maximum of the continuous function $f$ on $C$ exist because $C$ is compact, being the continuous image of $[0, 1]$. But $C$ is also connected for the same reason, so the range of $f$ must be an interval. The only option is that the range of $f$ must be the interval $[\min(f), \max(f)]$.
This is important because this number $\cfrac{1}{L_C} \int_C f(x_1, \ldots x_n)$ was just shown to be in that interval, and therefore is in the range of $f$. So, pick a value $\vec{c} \in C$ with $f(\vec{c}) = \cfrac{1}{L_C} \int_C f(x_1, \ldots x_n)$ and the theorem is proved.
EDIT: The added part of your question about the integral of a gradient vector field $f(x) = \nabla F(x)$ will have a negative answer. For example, a line integral of a gradient across a loop (initial point = endpoint) will give you zero, but the gradient vector field you are integrating might never be zero.
Example: Integrate the gradient of $F(x, y) = x + y$ along the counterclockwise oriented unit circle.
Presumably you mean that $f$ is monotone increasing for this to be true, and you are given no hypothesis regarding the existence of $f'$.
Of course, a monotone increasing function is differentiable almost everywhere and the derivative is integrable. However, we also would need to know that
$$\int_a^b f'(x) \, dx = f(b) - f(a)$$
for your approach, and that may not always be the case without additional assumptions, e.g., $f$ absolutely continuous.
Fortunately, no information about $f'$ is really needed to prove this. One approach is to use Riemann sums. This is straightforward but tedious.
Another way is to use Riemann-Stieltjes integrals and partial integration.
Defining $G(x) = \int_a^x g(t) \, dt$, we have since G(a) = 0,
$$\int_a^b f(x) g(x) \, dx = \int_a^b f \, dG = f(b)G(b)- f(a) G(a) - \int_a^b G \, df \\ = f(b)G(b) - \int_a^b G \, df $$
Since $G$ is continuous, by the first mean value theorem for integrals there exists $\theta$ such that
$$\int_a^b G \, df = G(\theta)\int_a^b df = G(\theta)(f(b) - f(a))$$
Thus, since $f(a) = 0$ we have
$$\int_a^b f(x) g(x) \, dx = f(b)G(b) - f(b)G(\theta) = f(b) \int_\theta^b g(x) \, dx$$
Best Answer
Some notations. First, for every $x$ in $[a,b]$, let $F(x)=\displaystyle\int_a^x\phi(t)\mathrm{d}t$, hence $F$ is continuous and $F(a)=0$. Second, assume that $G$ is nondecreasing, the other case being similar. Thus there exists a nonnegative measure $\mu$ such that, for every $x$ in $[a,b]$, $G(x)=G(a)+\displaystyle\int_a^x\mathrm{d}\mu(t)$.
Now to the proof. Using an integration by parts, the integral $I$ of $G\phi$ over $[a,b]$ is $$ I=[F(t)G(t)]_a^b-\int_a^bF(t)\mathrm{d}\mu(t)=F(b)G(b)-\int_a^bF(t)\mathrm{d}\mu(t). $$ The hypothesis made on $G$ means that $\mu$ is a nonnegative measure hence the first mean value theorem for integration yields the existence of a point $x$ in $[a,b]$ such that $$ \int_a^bF(t)\mathrm{d}\mu(t)=F(x)\int_a^b\mathrm{d}\mu(t)=F(x)(G(b)-G(a)). $$ Coming back to $I$, one gets $$ I=F(b)G(b)-F(x)(G(b)-G(a))=G(a)F(x)+G(b)(F(b)-F(x)), $$ which, by definition of $F$, is the desired assertion.
(No continuity of $G$ is required.)
In case you are wondering, the first mean value theorem used above is not the usual one but it has the same proof than the usual one. Namely, $F$ being continuous on the compact set $[a,b]$ has a maximum $M$ and a minimum $m$ on $[a,b]$, thus $$ m\int_a^b\mathrm{d}\mu(t)\le\int_a^bF(t)\mathrm{d}\mu(t)\le M\int_a^b\mathrm{d}\mu(t). $$ (This step uses the hypothesis that $\mu$ has constant sign.) In other words, there exists $u$ in $[m,M]$ such that $$ \int_a^bF(t)\mathrm{d}\mu(t)=u\int_a^b\mathrm{d}\mu(t). $$ Now, the continuity of $F$ implies that there exists $x$ in $[a,b]$ such that $u=F(x)$ and you are done.
Previous version When $\phi$ has a constant sign, this is a consequence of the intermediate value theorem. (An earlier version of this post did not assume $\phi$ of constant sign, hence an argument was wrong. Thanks to the OP for having asked some explanations.)
Let $I$ denote the integral of $G\phi$ over $[a,b]$, $J$ the integral of $\phi$ over $[a,b]$, and $H(x)$ the RHS of the equality you want to prove. Assume that $\phi$ is nonnegative and $G$ is nondecreasing, the other cases being similar.
Since $G(a)\le G(t)\le G(b)$ and $\phi(t)\ge0$ for every $t$ in $[a,b]$, $G(a)\phi(t)\le G(t)\phi(t)\le G(b)\phi(t)$ for every $t$, hence $G(a)J\le I\le G(b)J$. The function $x\mapsto H(x)$ is continuous on the interval $[a,b]$, $H(a)=G(b)J\ge I$ and $H(b)=G(a)J\le I$, hence by the intermediate value theorem there exists $x$ in $[a,b]$ such that $H(x)=I$.
(No continuity of $G$ is required.)