Morera's theorem, when properly(1) stated, is indeed the exact converse of Goursat's theorem.
Theorem (Morera): Let $\Omega\subset\mathbb{C}$ open, and $f\colon\Omega\to\mathbb{C}$ a continuous function. If for all triangles $T\subset\Omega$ whose interior is also contained in $\Omega$ $$\int_T f(z)\,dz = 0,$$ then $f$ is holomorphic in $\Omega$.
Instead of triangles, one could of course also use rectangles, or other polygons. And actually, we could drop the condition that the interior of the triangle be contained in $\Omega$ and be left with a still true, but arguably less useful result, since we would then have a sufficient but not necessary condition (consider $1/z$ on $\mathbb{C}\setminus \{0\}$ to have function satisfying the condition as stated, but not the stronger condition one obtains by dropping "whose interior is also contained in $\Omega$").
The condition entails the existence of local primitives of $f$, i.e. every $z\in \Omega$ has a neighbourhood $U$ such that $f = F'$ for a holomorphic function $F$ on $U$. Thus $f$ is holomorphic on $U$ (the derivative of a holomorphic function is again holomorphic), and since holomorphicity is a local property, $f$ is holomorphic on $\Omega$.
To establish the existence of local primitives, one considers (for example) for $z_0 \in \Omega$ a disk $U = D_r(z_0) \subset \Omega$, and on $U$ the function $F(z) = \int_{z_0}^z f(\zeta)\,d\zeta$. The vanishing of the integral of $f$ over triangles whose interior is contained in $\Omega$ then yields $F(z) - F(w) = \int_w^z f(\zeta)\,d\zeta$, from which $F' = f$ follows with the continuity of $f$.
(1) The term "properly" means "properly for this purpose", or "adequately to show it is the converse of Goursat's theorem" here. Stating it for simply connected domains or disks is not wrong.
There is one downside to stating it explicitly for simply connected domains, however. Often, people aren't aware of the local character of the theorem, and consider the simple connectedness as essential for the validity of the theorem. The essential point is the locality, that one considers not the entire domain $\Omega$, but a small convex neighbourhood $U\subset \Omega$ of a point $z\in \Omega$ to construct the local primitive.
No, it's not true.
Consider the function $f(z)=1/z$, which is holomorphic in the open set $\mathbb{C}\setminus\{0\}$, but has no antiderivative.
The key fact is that an open disc is simply connected, but the union of open discs may not be.
The statement is true for simply connected open sets, so it's true that you can find an antiderivative over an open disc around each point, but these may not “glue together”.
On a non simply connected open set there may exist functions not admitting an antiderivative.
Your idea cannot work: consider the situation above: there is no “large enough disc” that covers the given open set. Even if you consider $1/z$ over $D\setminus\{0\}$, where $D$ is the open disc at the origin with radius $1$, you incur in the same problem: whatever point $z_0$ you take in $D\setminus\{0\}$ you have a suitable disc around $z_0$ over which $f(z)=1/z$ has an antiderivative, but this disc can have radius at most $|z_0|$, because it cannot contain $0$.
What you could do is “glueing together” the local antiderivatives, but this doesn't work either. But it's the idea around analytic continuation.
Best Answer
Take an open disc around $z_0$. Then you can find $\varepsilon>0$ such that the closed path $$ \gamma: \phi\in[0,2\pi] \mapsto z_0 + \varepsilon e^{i\phi}$$ is inside the disc.
Now note that $$\oint_\gamma \bar z dz = i \epsilon\int_0^{2\pi} (\bar z_0 + \epsilon e^{-i\phi}) e^{i\phi} d\phi = i \epsilon^2 \int_0^{2\pi} d\phi = 2\pi i \epsilon^2 \ne0.$$ However, for $\bar z$ to have a primitive this integral should be $0$ as the path is closed.