More generally, you can say this: Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a continuous function. Let $\alpha:[0,1]\rightarrow\mathbb{R}^n$ and $\beta:[0,1]\rightarrow\mathbb{R}^n$ be two smooth curves. Suppose there is a smooth curve $\theta:[0,1]\rightarrow[0,1]$ such that $\theta(0)=0, \theta(1)=1$, and $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$. Then:
$$ \int_0^1 f(\alpha(t))\cdot \alpha'(t)dt = \int_0^1 f(\beta(t))\cdot \beta'(t)dt $$
Proof: By the chain rule we have $\beta'(t) = \alpha'(\theta(t))\theta'(t)$. Thus:
\begin{align}
\int_0^1 f(\beta(t))\cdot \beta'(t) dt &= \int_0^1 f(\alpha(\theta(t)))\cdot \alpha'(\theta(t))\theta'(t)dt \\
&= \int_0^1 f(\alpha(u))\cdot \alpha'(u)du
\end{align}
where the second equality is by the change of variables rule for integration of real-valued functions. $\Box$
In your example of a loop traversed once by a curve $\alpha$ and then twice by a curve $\beta$, you cannot write $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$. Defining $\theta(t) = 2t$ does not work because that would only be defined for $t \in [0,1/2]$. In another example of a clockwise loop versus a counter-clockwise loop, you again cannot find such a $\theta(t)$ function.
Edit: Some additional observations:
Allowing for non-injectivity:
The condition on $\theta$ above does not require the functions $\alpha, \beta$, or $\theta$ to be injective. For example, it works for the following: Let $\alpha(t)$ be any smooth curve defined over $t \in [0,1]$ (possibly having repeated values over that interval) and define $\beta(t) = \alpha(\theta(t))$, where
$$ \theta(t) = \sin^2\left(\frac{5\pi t}{2}\right) $$
The $\beta$ curve retraces the $\alpha$ curve 5 times, and the first four times cancel each other out.
Path independence when $f$ is a gradient
Suppose there is a smooth function $H:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $f(x) = \nabla H(x)$ for all $x \in \mathbb{R}^n$. Then path integrals of $f$ depend only on the endpoints of the path, not on the intermediate values of the path. Indeed:
$$ \int_0^1 f(\alpha(t))\cdot \alpha'(t) dt = \int_0^1\frac{d}{dt}\left[H(\alpha(t))\right] dt = H(\alpha(1))-H(\alpha(0)) $$
Cauchy integral formula
There are also connections to complex numbers and the Cauchy integral formula. For example, in some cases the path integral of a loopy curve depends only on the number of times the curve winds clockwise around the origin.
Best Answer
As the curve is contained in the sphere, the squared length $||\varphi(t)||^2 = \varphi(t)\cdot \varphi(t)$ is constantly equal to $R^2$. Differentiating with respect to $t$ gives $2\varphi(t)\cdot\dot{\varphi}(t) = 0$.